Trigonometry Class 9th Mathematics Term 2 Tamilnadu Board Solution

Class 9^th Mathematics Term 2 Tamilnadu Board Solution

Exercise 2.1

1. From the following diagrams, find the trigonometric ratios of the angle θ…
2. From the following diagrams, find the trigonometric ratios of the angle θ…
3. From the following diagrams, find the trigonometric ratios of the angle θ…
4. From the following diagrams, find the trigonometric ratios of the angle θ…
5. From the following diagrams, find the trigonometric ratios of the angle θ…
6. From the following diagrams, find the trigonometric ratios of the angle θ…
7. From the following diagrams, find the trigonometric ratios of the angle θ…
8. From the following diagrams, find the trigonometric ratios of the angle θ…
9. sina = 9/15 Find the other trigonometric ratios of the following
10. sina = 9/15 Find the other trigonometric ratios of the following
11. cosa = 15/17 Find the other trigonometric ratios of the following…
12. cosa = 15/17 Find the other trigonometric ratios of the following…
13. tanp = 5/12 Find the other trigonometric ratios of the following
14. tanp = 5/12 Find the other trigonometric ratios of the following
15. sectheta = 17/8 Find the other trigonometric ratios of the following…
16. sectheta = 17/8 Find the other trigonometric ratios of the following…
17. cosectheta = 61/60 Find the other trigonometric ratios of the following…
18. cosectheta = 61/60 Find the other trigonometric ratios of the following…
19. sintegrate heta = x/y Find the other trigonometric ratios of the following…
20. sintegrate heta = x/y Find the other trigonometric ratios of the following…
21. (i) sintegrate heta = 1/root 2 (ii) sin θ = 0 (iii) tantheta = root 3 (iv)…
22. (i) sintegrate heta = 1/root 2 (ii) sin θ = 0 (iii) tantheta = root 3 (iv)…
23. In ΔABC, right angled at B, AB = 10 and AC = 26. Find the six trigonometric…
24. In ΔABC, right angled at B, AB = 10 and AC = 26. Find the six trigonometric…
25. If 5cos θ - 12 sin θ = 0, find sintegrate heta +costheta /2costheta -sintegrate…
26. If 5cos θ - 12 sin θ = 0, find sintegrate heta +costheta /2costheta -sintegrate…
27. If 29cosθ = 20 find sec^2 θ - tan^2 θ.
28. If 29cosθ = 20 find sec^2 θ - tan^2 θ.
29. If sectheta = 26/10 find 3costheta +4sintegrate heta /4costheta -2sintegrate…
30. If sectheta = 26/10 find 3costheta +4sintegrate heta /4costheta -2sintegrate…
31. If tantheta = a/b find sin^2 θ + cos^2 θ.
32. If tantheta = a/b find sin^2 θ + cos^2 θ.
33. If cottheta = 15/8 , evaluate (1+sintegrate heta) (1-sintegrate
34. If cottheta = 15/8 , evaluate (1+sintegrate heta) (1-sintegrate
35. In triangle PQR, right angled at Q, if tan p = 1/root 3 find the value of (i)…
36. In triangle PQR, right angled at Q, if tan p = 1/root 3 find the value of (i)…
37. If sectheta = 13/5 show that 2sintegrate heta -3costheta /4sintegrate heta…
38. If sectheta = 13/5 show that 2sintegrate heta -3costheta /4sintegrate heta…
39. If seca = 17/8 prove that 1-2 sin^2 A = 2cos^2 A - 1.
40. If seca = 17/8 prove that 1-2 sin^2 A = 2cos^2 A - 1.
41. (i) sin 45° + cos 45° (ii) sin 60° tan 30° (iii) tan45^circle /tan30^circle +…
42. (i) sin 45° + cos 45° (ii) sin 60° tan 30° (iii) tan45^circle /tan30^circle +…
43. (i) sin^2 30° + cos^2 30° = 1 (ii) 1 + tan^2 45° = sec^2 45° (iii) cos 60° = 1…
44. (i) sin^2 30° + cos^2 30° = 1 (ii) 1 + tan^2 45° = sec^2 45° (iii) cos 60° = 1…

Exercise 2.2

1. sin36^circle /cos54^circle Evaluate
2. sin36^circle /cos54^circle Evaluate
3. cosec10^circle /sec80^circle Evaluate
4. cosec10^circle /sec80^circle Evaluate
5. sin θ sec(90° - θ) Evaluate
6. sin θ sec(90° - θ) Evaluate
7. sec20^circle /cosec70^circle Evaluate
8. sec20^circle /cosec70^circle Evaluate
9. sin17^circle /cos73^circle Evaluate
10. sin17^circle /cos73^circle Evaluate
11. tan46^circle /cot44^circle Evaluate
12. tan46^circle /cot44^circle Evaluate
13. cos 38° cos 52° - sin 38° sin 52° Simplify
14. cos 38° cos 52° - sin 38° sin 52° Simplify
15. cos80^circle /sin10^circle + cos59^circle cosec31^circle Simplify…
16. cos80^circle /sin10^circle + cos59^circle cosec31^circle Simplify…
17. sin36^circle /cos54^circle - tan54^circle /cot36^circle Simplify
18. sin36^circle /cos54^circle - tan54^circle /cot36^circle Simplify
19. 3 tan67^circle /cot23^circle + 1/2 sin42^circle /cos48^circle + 5/2…
20. 3 tan67^circle /cot23^circle + 1/2 sin42^circle /cos48^circle + 5/2…
21. cos37^circle /sin53^circle x sin18^circle /cos72^circle Simplify
22. cos37^circle /sin53^circle x sin18^circle /cos72^circle Simplify
23. 2 sec (90^circle - theta)/cosectheta +7 cos (90^circle - theta)/sintegrate heta…
24. 2 sec (90^circle - theta)/cosectheta +7 cos (90^circle - theta)/sintegrate heta…
25. sec (90^circle - theta)/sin (90^circle - theta) x costheta /tan (90^circle -…
26. sec (90^circle - theta)/sin (90^circle - theta) x costheta /tan (90^circle -…
27. sin35^circle /cos55^circle + cos55^circle /sin35^circle - 2cos^260^circle…
28. sin35^circle /cos55^circle + cos55^circle /sin35^circle - 2cos^260^circle…
29. cot12° cot38° cot52° cot60° cot78°. Simplify
30. cot12° cot38° cot52° cot60° cot78°. Simplify
31. (i) sin A = cos 30° (ii) tan49° = cot A (iii) tan A tan 35° = 1 (iv) sec 35° =…
32. (i) sin A = cos 30° (ii) tan49° = cot A (iii) tan A tan 35° = 1 (iv) sec 35° =…
33. cos 48° - sin 42° = 0 Show that
34. cos 48° - sin 42° = 0 Show that
35. cos 20° cos 70° - sin 70° sin 20° = 0 Show that
36. cos 20° cos 70° - sin 70° sin 20° = 0 Show that
37. sin (90° - θ)tan θ = sin θ Show that
38. sin (90° - θ)tan θ = sin θ Show that
39. cos (90^circle - theta) tan (90^circle - theta)/costheta = 1 Show that…
40. cos (90^circle - theta) tan (90^circle - theta)/costheta = 1 Show that…

Exercise 2.3

1. sin 26o Find the value of the following.
2. sin 26o Find the value of the following.
3. cos 72o Find the value of the following.
4. cos 72o Find the value of the following.
5. tan35o Find the value of the following.
6. tan35o Find the value of the following.
7. sin 75o 15’ Find the value of the following.
8. sin 75o 15’ Find the value of the following.
9. sin 12° 12’ Find the value of the following.
10. sin 12° 12’ Find the value of the following.
11. cos 12o 35’ Find the value of the following.
12. cos 12o 35’ Find the value of the following.
13. cos 40o 20’ Find the value of the following.
14. cos 40o 20’ Find the value of the following.
15. tan 10o 26’ Find the value of the following.
16. tan 10o 26’ Find the value of the following.
17. cot 20o Find the value of the following.
18. cot 20o Find the value of the following.
19. cot 40^0 20’ Find the value of the following.
20. cot 40^0 20’ Find the value of the following.
21. (i) sinθ= 0.7009 (ii) cos θ = 0.9664 (iii) tan θ = 0.3679 (iv) cotθ = 0.2334 (v)…
22. (i) sinθ= 0.7009 (ii) cos θ = 0.9664 (iii) tan θ = 0.3679 (iv) cotθ = 0.2334 (v)…
23. sin 30°30’+cos 40°20’ Simplify, using trigonometric tables
24. sin 30°30’+cos 40°20’ Simplify, using trigonometric tables
25. tan 45° 27’ + sin 20° Simplify, using trigonometric tables
26. tan 45° 27’ + sin 20° Simplify, using trigonometric tables
27. tan 63°12’ - cos 12°42’ Simplify, using trigonometric tables
28. tan 63°12’ - cos 12°42’ Simplify, using trigonometric tables
29. sin 50° 26’ + cos 18° + tan 70° 12’ Simplify, using trigonometric tables…
30. sin 50° 26’ + cos 18° + tan 70° 12’ Simplify, using trigonometric tables…
31. tan 72° + cot 30° Simplify, using trigonometric tables
32. tan 72° + cot 30° Simplify, using trigonometric tables
33. Find the area of the right triangle with hypotenuse 20cm and one of the acute…
34. Find the area of the right triangle with hypotenuse 20cm and one of the acute…
35. Find the area of the triangle with hypotenuse 8cm and one of the acute angle is…
36. Find the area of the triangle with hypotenuse 8cm and one of the acute angle is…
37. Find the area of isosceles triangle with base 16cm and vertical angle 60° 40’…
38. Find the area of isosceles triangle with base 16cm and vertical angle 60° 40’…
39. Find the area of isosceles triangle with base 15cm and vertical angle 80°…
40. Find the area of isosceles triangle with base 15cm and vertical angle 80°…
41. A ladder makes an angle 30° with the floor and its lower end is 12m away from…
42. A ladder makes an angle 30° with the floor and its lower end is 12m away from…
43. Find the angle made by a ladder of length 4m with the ground if its one end is…
44. Find the angle made by a ladder of length 4m with the ground if its one end is…
45. Find the length of the chord of a circle of radius 5cm subtending an angle of…
46. Find the length of the chord of a circle of radius 5cm subtending an angle of…
47. Find the length of the side of regular polygon of 12 sides inscribed in a…
48. Find the length of the side of regular polygon of 12 sides inscribed in a…
49. Find the radius of the incircle of a regular hexagon of side 24cm.…
50. Find the radius of the incircle of a regular hexagon of side 24cm.…

Exercise 2.4

1. The value of sin^2 60° + cos^2 60° is equal toA. sin^2 45° + cos^2 45° B. tan^2…
2. The value of sin^2 60° + cos^2 60° is equal toA. sin^2 45° + cos^2 45° B. tan^2…
3. If x = 2tan30^circle /1-tan^230^circle , then the value of x isA. tan 45° B. tan…
4. If x = 2tan30^circle /1-tan^230^circle , then the value of x isA. tan 45° B. tan…
5. The value of sec^2 45° - tan^2 45° is equal toA. sin^2 60° - cos^2 60° B. sin^2…
6. The value of sec^2 45° - tan^2 45° is equal toA. sin^2 60° - cos^2 60° B. sin^2…
7. The value of 2sin30° cos30° is equal toA. tan 30° B. cos 60° C. sin 60° D. cot…
8. The value of 2sin30° cos30° is equal toA. tan 30° B. cos 60° C. sin 60° D. cot…
9. The value of cosec^2 60° - 1 is equal toA. cos^2 60° B. cot^2 60° C. sec^2 60°…
10. The value of cosec^2 60° - 1 is equal toA. cos^2 60° B. cot^2 60° C. sec^2 60°…
11. cos60° cos30° - sin60° sin30° is equal toA. cos 90° B. cosec 90° C. sin 30° +…
12. cos60° cos30° - sin60° sin30° is equal toA. cos 90° B. cosec 90° C. sin 30° +…
13. The value of sin27^0/cos63^circle isA.0 B. 1 C. tan 27° D. cot 63°…
14. The value of sin27^0/cos63^circle isA.0 B. 1 C. tan 27° D. cot 63°…
15. If cos x = sin 43°, then the value of x isA. 57° B. 43° C. 47° D. 90°…
16. If cos x = sin 43°, then the value of x isA. 57° B. 43° C. 47° D. 90°…
17. The value of sec 29° - cosec 61° isA. 1 B. 0 C. sec 60° D. cosec 29°…
18. The value of sec 29° - cosec 61° isA. 1 B. 0 C. sec 60° D. cosec 29°…
19. If 3x cosec 36° = sec 54°, then the value of x isA. 0 B. 1 C. 1/3 D. 3/4…
20. If 3x cosec 36° = sec 54°, then the value of x isA. 0 B. 1 C. 1/3 D. 3/4…
21. The value of sin60° cos30° + cos60° sin30° is equal toA. sec 90° B. tan 90° C.…
22. The value of sin60° cos30° + cos60° sin30° is equal toA. sec 90° B. tan 90° C.…
23. If cosacos30^circle = root 3/4 ,then the measure of A isA. 90° B. 60° C. 45° D.…
24. If cosacos30^circle = root 3/4 ,then the measure of A isA. 90° B. 60° C. 45° D.…
25. The value of tan26^circle /cot64^circle isA. 1/2 B. root 3/2 C. 0 D. 1…
26. The value of tan26^circle /cot64^circle isA. 1/2 B. root 3/2 C. 0 D. 1…
27. The value of sin 60° - cos 30° isA. 0 B. 1/root 2 C. root 3/2 D. 1…
28. The value of sin 60° - cos 30° isA. 0 B. 1/root 2 C. root 3/2 D. 1…
29. The value of cos^2 30° - sin^2 30° isA. cos 60° B. sin 60° C. 0 D. 1…
30. The value of cos^2 30° - sin^2 30° isA. cos 60° B. sin 60° C. 0 D. 1…

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Exercise 2.1

Question 1.

From the following diagrams, find the trigonometric ratios of the angle θ

Answer:

⇒ sin θ = 

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 2.

From the following diagrams, find the trigonometric ratios of the angle θ

Answer:

⇒ sin θ = 

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 3.

From the following diagrams, find the trigonometric ratios of the angle θ

Answer:

sin θ = 

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 4.

From the following diagrams, find the trigonometric ratios of the angle θ

Answer:

sin θ = 

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 5.

From the following diagrams, find the trigonometric ratios of the angle θ

Answer:

sin θ = 

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 6.

From the following diagrams, find the trigonometric ratios of the angle θ

Answer:

sin θ = 

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 7.

From the following diagrams, find the trigonometric ratios of the angle θ

Answer:

sin θ = 

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 8.

From the following diagrams, find the trigonometric ratios of the angle θ

Answer:

sin θ = 

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 9.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 9² + p² = 15²

⇒ 81 + p² = 225

⇒ p² = 144

⇒ p = 12

Therefore, the other angles are:

cos A = 

tan A = 

cosec A = 

sec A = 

cot A = 

Question 10.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 9² + p² = 15²

⇒ 81 + p² = 225

⇒ p² = 144

⇒ p = 12

Therefore, the other angles are:

cos A = 

tan A = 

cosec A = 

sec A = 

cot A = 

Question 11.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 15² + p² = 17²

⇒ 225 + p² = 289

⇒ p² = 64

⇒ p = 8

Therefore, the other angles are:

sin P = 

tan P = 

cosec P = 

sec P = 

cot P = 

Question 12.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 15² + p² = 17²

⇒ 225 + p² = 289

⇒ p² = 64

⇒ p = 8

Therefore, the other angles are:

sin P = 

tan P = 

cosec P = 

sec P = 

cot P = 

Question 13.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + 12² = P²

⇒ 25 + 144 = p²

⇒ p² = 169

⇒ p = 13

Therefore, the other angles are:

sin P = 

cos P = 

cosec P = 

sec P = 

cot P = 

Question 14.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + 12² = P²

⇒ 25 + 144 = p²

⇒ p² = 169

⇒ p = 13

Therefore, the other angles are:

sin P = 

cos P = 

cosec P = 

sec P = 

cot P = 

Question 15.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 15² + p² = 17²

⇒ 64 + p² = 289

⇒ p² = 225

⇒ p = 15

Therefore, the other angles are:

sin θ = 

cos θ = 

tan θ = 

cosec θ = 

cot θ = 

Question 16.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 15² + p² = 17²

⇒ 64 + p² = 289

⇒ p² = 225

⇒ p = 15

Therefore, the other angles are:

sin θ = 

cos θ = 

tan θ = 

cosec θ = 

cot θ = 

Question 17.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 60² + p² = 61²

⇒ 3600 + p² = 3721

⇒ p² = 121

⇒ p = 11

Therefore, the other angles are:

sin θ = 

cos θ = 

tan θ = 

sec θ = 

cot θ = 

Question 18.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 60² + p² = 61²

⇒ 3600 + p² = 3721

⇒ p² = 121

⇒ p = 11

Therefore, the other angles are:

sin θ = 

cos θ = 

tan θ = 

sec θ = 

cot θ = 

Question 19.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ x² + p² = y²

⇒ p² = y² – x²

⇒ p = √y² – x²

Therefore, the other angles are:

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 20.

Find the other trigonometric ratios of the following

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ x² + p² = y²

⇒ p² = y² – x²

⇒ p = √y² – x²

Therefore, the other angles are:

cos θ = 

tan θ = 

cosec θ = 

sec θ = 

cot θ = 

Question 21.

Find the value of θ, if

(i) 

(ii) sin θ = 0

(iii) 

(iv) 

Answer:

(i) 

θ = 45° (From the table)

(ii) 

θ = 0° (From the table)

(iii) 

θ = 60° (From the table)

(iv) 

θ = 30° (From the table)

Question 22.

Find the value of θ, if

(i) 

(ii) sin θ = 0

(iii) 

(iv) 

Answer:

(i) 

sin θ = sin 45°

θ = 45° (From the table)

(ii) 

sin θ = sin 0° 

θ = 0° (From the table)

(iii) 

tan θ = tan 60°

θ = 60° (From the table)

(iv) 

cos θ = cos 30° 

θ = 30° (From the table)

Question 23.

In ΔABC, right angled at B, AB = 10 and AC = 26. Find the six trigonometric ratios of the angles A and C.

Answer:

For ∠A:

sin A = 

cos A = 

tan A = 

cosec A = 

sec A = 

cot A = 

For ∠C:

sin C = 

cos C = 

tan C = 

cosec C = 

sec C = 

cot C = 

Question 24.

In ΔABC, right angled at B, AB = 10 and AC = 26. Find the six trigonometric ratios of the angles A and C.

Answer:

For ∠A:

sin A = 

cos A = 

tan A = 

cosec A = 

sec A = 

cot A = 

For ∠C:

sin C = 

cos C = 

tan C = 

cosec C = 

sec C = 

cot C = 

Question 25.

If 5cos θ – 12 sin θ = 0, find 

Answer:

5cosθ –12sinθ = 0

⇒ 5cosθ = 12sinθ

⇒ tanθ = 5/12

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + 12² = P²

⇒ 25 + 144 = p²

⇒ p² = 169

⇒ p = 13

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting values in the function as:

⇒ 

⇒ 

⇒ 

Question 26.

If 5cos θ – 12 sin θ = 0, find 

Answer:

5cosθ –12sinθ = 0

⇒ 5cosθ = 12sinθ

⇒ tanθ = 5/12

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + 12² = P²

⇒ 25 + 144 = p²

⇒ p² = 169

⇒ p = 13

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting values in the function as:

⇒ 

⇒ 

⇒ 

Question 27.

If 29cosθ = 20 find sec² θ – tan² θ.

Answer:

29cosθ = 20

⇒ cosθ = 20/29

Let the third side be p,

By Pythagoras theorem,

⇒ 20² + p² = 29²

⇒ 400 + p² = 841

⇒ p² = 441

⇒ p = 21

Therefore, the other angles are:

sec θ = 

tan θ = 

Putting the values in function:

⇒ sec² θ – tan² θ

⇒ 

⇒ 

⇒ 

Question 28.

If 29cosθ = 20 find sec² θ – tan² θ.

Answer:

29cosθ = 20

⇒ cosθ = 20/29

Let the third side be p,

By Pythagoras theorem,

⇒ 20² + p² = 29²

⇒ 400 + p² = 841

⇒ p² = 441

⇒ p = 21

Therefore, the other angles are:

sec θ = 

tan θ = 

Putting the values in function:

⇒ sec² θ – tan² θ

⇒ 

⇒ 

⇒ 

Question 29.

If  find .

Answer:

∠BAC = θ

Let the third side be p,

By Pythagoras theorem,

⇒ 10² + p² = 26²

⇒ 100 + p² = 676

⇒ p² = 576

⇒ p = 24

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting values in the function as:

⇒ 

⇒ 

⇒ 

Question 30.

If  find .

Answer:

∠BAC = θ

Let the third side be p,

By Pythagoras theorem,

⇒ 10² + p² = 26²

⇒ 100 + p² = 676

⇒ p² = 576

⇒ p = 24

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting values in the function as:

⇒ 

⇒ 

⇒ 

Question 31.

If  find sin² θ + cos² θ.

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ a² + b² = p²

⇒ p = √a² + b²

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting the values in function:

⇒ sin² θ + cos² θ

⇒ 

⇒ 

⇒ 

Question 32.

If  find sin² θ + cos² θ.

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ a² + b² = p²

⇒ p = √a² + b²

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting the values in function:

⇒ sin² θ + cos² θ

⇒ 

⇒ 

⇒ 

Question 33.

If , evaluate

Answer:

Simplifying the function:

⇒ 

⇒ 

Let the third side be p,

By Pythagoras theorem,

⇒ 8² + 15² = P²

⇒ 64 + 225 = p²

⇒ p² = 289

⇒ p = 17

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting these values in the simplied function:
⇒ 

⇒ 

⇒ 

⇒ 

Question 34.

If , evaluate

Answer:

Simplifying the function:

⇒ 

⇒ 

Let the third side be p,

By Pythagoras theorem,

⇒ 8² + 15² = P²

⇒ 64 + 225 = p²

⇒ p² = 289

⇒ p = 17

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting these values in the simplied function:
⇒ 

⇒ 

⇒ 

⇒ 

Question 35.

In triangle PQR, right angled at Q, if tan find the value of

(i) sin P cos R + cos P sin R

(ii) cos P cos R – sin P sin R.

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 1² + (√3)² = P²

⇒ 1 + 3 = p²

⇒ p² = 4

⇒ p = 2

Therefore, the other angles are:

sin P = 

cos P = 

sin R = 

cos R = 

(i) Putting the values in the expression:

sin P cos R + cos P sin R

⇒ 

⇒ 

⇒ 1

(ii) Putting the values in the expression:

cos P cos R – sin P sin R.

⇒ 

⇒ 0

Question 36.

In triangle PQR, right angled at Q, if tan find the value of

(i) sin P cos R + cos P sin R

(ii) cos P cos R – sin P sin R.

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 1² + (√3)² = P²

⇒ 1 + 3 = p²

⇒ p² = 4

⇒ p = 2

Therefore, the other angles are:

sin P = 

cos P = 

sin R = 

cos R = 

(i) Putting the values in the expression:

sin P cos R + cos P sin R

⇒ 

⇒ 

⇒ 1

(ii) Putting the values in the expression:

cos P cos R – sin P sin R.

⇒ 

⇒ 0

Question 37.

If show that 

Answer: 

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + p² = 13²

⇒ 25 + p² = 169

⇒ p² = 144

⇒ p = 12

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting these values in the left hand side:
⇒ 

⇒ 

⇒ 

⇒ 3

Which is equal to right hand side.

Hence proved.

Question 38.

If show that 

Answer: 

Let the third side be p,

By Pythagoras theorem,

⇒ 5² + p² = 13²

⇒ 25 + p² = 169

⇒ p² = 144

⇒ p = 12

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting these values in the left hand side:
⇒ 

⇒ 

⇒ 

⇒ 3

Which is equal to right hand side.

Hence proved.

Question 39.

If prove that 1–2 sin²A = 2cos²A – 1.

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 8² + p² = 17²

⇒ 64 + p² = 289

⇒ p² = 225

⇒ p = 15

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting these values in the left hand side:
⇒ 1–2×

⇒ 1–2×

⇒ 1–

⇒ 

Putting values in the right hand side:

⇒ 2×

⇒ 2× –1

⇒ 

⇒ 

Therefore, Left hand side and right hand side are equal.

Hence proved.

Question 40.

If prove that 1–2 sin²A = 2cos²A – 1.

Answer:

Let the third side be p,

By Pythagoras theorem,

⇒ 8² + p² = 17²

⇒ 64 + p² = 289

⇒ p² = 225

⇒ p = 15

Therefore, the other angles are:

sin θ = 

cos θ = 

Putting these values in the left hand side:
⇒ 1–2×

⇒ 1–2×

⇒ 1–

⇒ 

Putting values in the right hand side:

⇒ 2×

⇒ 2× –1

⇒ 

⇒ 

Therefore, Left hand side and right hand side are equal.

Hence proved.

Question 41.

Evaluate.

(i) sin 45° + cos 45°

(ii) sin 60° tan 30°

(iii) 

(iv) cos²60° sin²30° + tan² 30° cot² 60°

(v) 6cos² 90° + 3sin² 90° + 4 tan² 45°

(vi) 

(vii) 

(viii) 4(sin⁴ 30° + cos⁴ 60°) – 3 (cos² 45° – sin² 90°)

Answer:

(i) We know sin 45° = (1/√2) and cos 45° = (1/√2) (From the table)

⇒ 

⇒ 

(ii) We know sin 30° = 1/2 and tan 45° = 1 (From the table)

⇒ 

(iii) We know tan 30° = (1/√3) (From the table)

tan 45° = 1

tan 60° = √3

⇒ 

⇒ 

⇒ 

(iv) We know sin 30° = cos 60° = 1/2 (From the table)

tan 30° = cot 60° = 1/√3

⇒ 

⇒ 

⇒ 

⇒ 

(v) We know that cos 90° = 0 , sin 90° = 1 and tan 45° = 1 (From the table)

⇒ 6(0)² + 3(1)² + 4(1)²

⇒ 3 + 4

⇒ 7

(vi) We know cot 60° = (1/√3) (From the table)

sec 30° = 2/√3

sin 45° = 1/√2

sin 60° = √3/2

cos 45° = 1/√2

⇒ 

⇒ 

⇒ 

⇒ 

(vii) We know tan 60° = √3 (From the table)

cos 45° = 1/√2

sec 30° = 2/√3

cos 90° = 0

cosec 30° = 2

sec 60° = 2

cot 30° = √3

⇒ 

⇒ 

⇒ 9

(viii) We know

sin 30° = cos 60° = 1/2 (From the table)

cos 45° = 1/√2

sin 90° = 1

⇒ 4(

⇒ 4(

⇒ 4(

⇒ 2

Question 42.

Evaluate.

(i) sin 45° + cos 45°

(ii) sin 60° tan 30°

(iii) 

(iv) cos²60° sin²30° + tan² 30° cot² 60°

(v) 6cos² 90° + 3sin² 90° + 4 tan² 45°

(vi) 

(vii) 

(viii) 4(sin⁴ 30° + cos⁴ 60°) – 3 (cos² 45° – sin² 90°)

Answer:

(i) We know sin 45° = (1/√2) and cos 45° = (1/√2) (From the table)

⇒ 

⇒ 

(ii) We know sin 30° = 1/2 and tan 45° = 1 (From the table)

⇒ 

(iii) We know tan 30° = (1/√3) (From the table)

tan 45° = 1

tan 60° = √3

⇒ 

⇒ 

⇒ 

(iv) We know sin 30° = cos 60° = 1/2 (From the table)

tan 30° = cot 60° = 1/√3

⇒ 

⇒ 

⇒ 

⇒ 

(v) We know that cos 90° = 0 , sin 90° = 1 and tan 45° = 1 (From the table)

⇒ 6(0)² + 3(1)² + 4(1)²

⇒ 3 + 4

⇒ 7

(vi) We know cot 60° = (1/√3) (From the table)

sec 30° = 2/√3

sin 45° = 1/√2

sin 60° = √3/2

cos 45° = 1/√2

⇒ 

⇒ 

⇒ 

⇒ 

(vii) We know tan 60° = √3 (From the table)

cos 45° = 1/√2

sec 30° = 2/√3

cos 90° = 0

cosec 30° = 2

sec 60° = 2

cot 30° = √3

⇒ 

⇒ 

⇒ 9

(viii) We know

sin 30° = cos 60° = 1/2 (From the table)

cos 45° = 1/√2

sin 90° = 1

⇒ 4(

⇒ 4(

⇒ 4(

⇒ 2

Question 43.

Verify the following equalities.

(i) sin²30° + cos²30° = 1

(ii) 1 + tan²45° = sec²45°

(iii) cos 60° = 1 – 2sin² 30° = 2cos² 30° – 1

(iv) cos 90° = 1 – 2 sin² 45° = 2cos² 45° – 1

(v) 

(vi) 

(vii) 

(viii) tan²60° – 2tan²45° – cot²30° + 2sin²

(ix) 4cot²45° – sec²60° + sin² 60° = 1

(x) sin30° cos60° + cos30° sin60° = sin 90°

Answer:

(i) We know sin 30° = (1/2) and cos 30° = (√3/2) (From the table)

Putting the values in the left hand side:

⇒ 

⇒ 

⇒ 1

Which is equal to the right hand side.

Hence verified.

(ii) We know tan 45° = 1 and sec 45° = √2 (From the table)

Putting the values in the left hand side:

⇒ 1 + (1)²

⇒ 1 + 1

⇒ 2

Putting the values in the right hand side:

⇒ (√2)²

⇒ 2

∴ LHS = RHS

Hence verified.

(iii) We know sin 30° = cos 60° = (1/2) and cos 30° = (√3/2) (From the table)

So the leftmost function = cos60° = (1/2)

Putting values in the middle function:
⇒ 1 – 2× 

⇒ 

⇒ 1/2

Putting values in the rightmost function:
⇒ 2× 

⇒ 

⇒ 1/2

Therefore, all simplify to 1/2 and are equal.

Hence verified.

(iv) We know sin 45° = cos 45° = (1/√ 2) and cos 90° = 0 (From the table)

So the leftmost function = cos90° = 0

Putting values in the middle function:
⇒ 1 – 2× 

⇒ 

⇒ 0

Putting values in the rightmost function:
⇒ 2× 

⇒ 

⇒ 0

Therefore, all simplify to 0 and are equal.

Hence verified.

(v) We know sin 60° = (√3/2) (From the table)

cos 60° = (1/2)

tan 60° = √3

and sec 30° = 2

Putting values in the left hand side:
⇒ 

⇒ 

⇒ 

Putting values in the right hand side:
⇒ 

Therefore, left hand side and right hand side are equal.

Hence verified.

(vi) We know cos 60° = (1/2) (From the table)

tan 60° = √3

Putting values in the left hand side:
⇒ 

⇒ 

⇒ 

Putting values in the right hand side:
⇒ 

⇒ 

⇒ 

⇒ 

Therefore, left hand side and right hand side are equal.

Hence verified.

(vii) We know sec 60° = (2/√3) (From the table)

tan 30° = (1/√3)

sin 30° = (1/2)

Putting values in the left hand side:
⇒ 

⇒ 

⇒ 3

Putting values in the right hand side:
⇒ 

⇒ 

⇒ 3

Therefore, left hand side and right hand side are equal.

Hence verified.

(viii) We know sin 30° = (1/2) (From the table)

cot 30° = √3

cosec 45° = √2

tan 60° = √3

and tan 45° = 1

Putting the values in the left hand side:

⇒ 

⇒ 

⇒ 0

Which is equal to the right hand side.

Hence verified.

(ix) We know cos 60° = (1/2) (From the table)

sin 60° = (√3/2)

sec 60° = 2

and cot 45° = 1

Putting the values in the left hand side:

⇒ 

⇒ 

⇒ 1

Which is equal to the right hand side.

Hence verified.

(x) We know sin 30° = cos 60° = (1/2) (From the table)

Cos 30° = sin 60° = (√3/2)

Sin 90° = 1

So the right hand side = sin 90° = 1

Putting the values in the left hand side:

⇒ 

⇒ 

⇒ 1

Which is equal to the right hand side.

Hence verified.

Question 44.

Verify the following equalities.

(i) sin²30° + cos²30° = 1

(ii) 1 + tan²45° = sec²45°

(iii) cos 60° = 1 – 2sin² 30° = 2cos² 30° – 1

(iv) cos 90° = 1 – 2 sin² 45° = 2cos² 45° – 1

(v) 

(vi) 

(vii) 

(viii) tan²60° – 2tan²45° – cot²30° + 2sin²

(ix) 4cot²45° – sec²60° + sin² 60° = 1

(x) sin30° cos60° + cos30° sin60° = sin 90°

Answer:

(i) We know sin 30° = (1/2) and cos 30° = (√3/2) (From the table)

Putting the values in the left hand side:

⇒ 

⇒ 

⇒ 1

Which is equal to the right hand side.

Hence verified.

(ii) We know tan 45° = 1 and sec 45° = √2 (From the table)

Putting the values in the left hand side:

⇒ 1 + (1)²

⇒ 1 + 1

⇒ 2

Putting the values in the right hand side:

⇒ (√2)²

⇒ 2

∴ LHS = RHS

Hence verified.

(iii) We know sin 30° = cos 60° = (1/2) and cos 30° = (√3/2) (From the table)

So the leftmost function = cos60° = (1/2)

Putting values in the middle function:
⇒ 1 – 2× 

⇒ 

⇒ 1/2

Putting values in the rightmost function:
⇒ 2× 

⇒ 

⇒ 1/2

Therefore, all simplify to 1/2 and are equal.

Hence verified.

(iv) We know sin 45° = cos 45° = (1/√ 2) and cos 90° = 0 (From the table)

So the leftmost function = cos90° = 0

Putting values in the middle function:
⇒ 1 – 2× 

⇒ 

⇒ 0

Putting values in the rightmost function:
⇒ 2× 

⇒ 

⇒ 0

Therefore, all simplify to 0 and are equal.

Hence verified.

(v) We know sin 60° = (√3/2) (From the table)

cos 60° = (1/2)

tan 60° = √3

and sec 30° = 2

Putting values in the left hand side:
⇒ 

⇒ 

⇒ 

Putting values in the right hand side:
⇒ 

Therefore, left hand side and right hand side are equal.

Hence verified.

(vi) We know cos 60° = (1/2) (From the table)

tan 60° = √3

Putting values in the left hand side:
⇒ 

⇒ 

⇒ 

Putting values in the right hand side:
⇒ 

⇒ 

⇒ 

⇒ 

Therefore, left hand side and right hand side are equal.

Hence verified.

(vii) We know sec 60° = (2/√3) (From the table)

tan 30° = (1/√3)

sin 30° = (1/2)

Putting values in the left hand side:
⇒ 

⇒ 

⇒ 3

Putting values in the right hand side:
⇒ 

⇒ 

⇒ 3

Therefore, left hand side and right hand side are equal.

Hence verified.

(viii) We know sin 30° = (1/2) (From the table)

cot 30° = √3

cosec 45° = √2

tan 60° = √3

and tan 45° = 1

Putting the values in the left hand side:

⇒ 

⇒ 

⇒ 0

Which is equal to the right hand side.

Hence verified.

(ix) We know cos 60° = (1/2) (From the table)

sin 60° = (√3/2)

sec 60° = 2

and cot 45° = 1

Putting the values in the left hand side:

⇒ 

⇒ 

⇒ 1

Which is equal to the right hand side.

Hence verified.

(x) We know sin 30° = cos 60° = (1/2) (From the table)

Cos 30° = sin 60° = (√3/2)

Sin 90° = 1

So the right hand side = sin 90° = 1

Putting the values in the left hand side:

⇒ 

⇒ 

⇒ 1

Which is equal to the right hand side.

Hence verified.

---

Exercise 2.2

Question 1.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 2.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 3.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 4.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 5.

Evaluate

sin θ sec(90° – θ)

Answer:

⇒ sin θ sec (90–θ)

⇒ sin θ cosec θ

⇒ sin θ × (1/sin θ )

⇒ 1

Question 6.

Evaluate

sin θ sec(90° – θ)

Answer:

⇒ sin θ sec (90–θ)

⇒ sin θ cosec θ

⇒ sin θ × (1/sin θ )

⇒ 1

Question 7.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 8.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 9.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 10.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 11.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 12.

Evaluate

Answer:

⇒ 

⇒ 

⇒ 

⇒ 1

Question 13.

Simplify

cos 38° cos 52° – sin 38° sin 52°

Answer:

⇒ cos 38° cos 52° – sin 38° sin 52°

⇒ cos 38° sin (90–52)° – sin 38° cos (90–52)°

⇒ cos 38° sin 38° – sin 38° cos 38°

⇒ 0

Question 14.

Simplify

cos 38° cos 52° – sin 38° sin 52°

Answer:

⇒ cos 38° cos 52° – sin 38° sin 52°

⇒ cos 38° sin (90–52)° – sin 38° cos (90–52)°

⇒ cos 38° sin 38° – sin 38° cos 38°

⇒ 0

Question 15.

Simplify

Answer:

⇒ 

⇒ 

⇒ 1 + 1 = 2

Question 16.

Simplify

Answer:

⇒ 

⇒ 

⇒ 1 + 1 = 2

Question 17.

Simplify

Answer:

⇒ 

⇒ 

⇒ 1 – 1 = 0

Question 18.

Simplify

Answer:

⇒ 

⇒ 

⇒ 1 – 1 = 0

Question 19.

Simplify

Answer:

⇒ 

⇒ 

⇒ (3×1) + (1/2)×1 + (5/2)×1

⇒ 3 + 3 = 6

Question 20.

Simplify

Answer:

⇒ 

⇒ 

⇒ (3×1) + (1/2)×1 + (5/2)×1

⇒ 3 + 3 = 6

Question 21.

Simplify

Answer:

⇒ 

⇒ 

⇒ 1 × 1 = 1

Question 22.

Simplify

Answer:

⇒ 

⇒ 

⇒ 1 × 1 = 1

Question 23.

Simplify

Answer:

⇒ 

⇒ 2 × 1 + 7 × 1

⇒ 2 + 7 = 9

Question 24.

Simplify

Answer:

⇒ 

⇒ 2 × 1 + 7 × 1

⇒ 2 + 7 = 9

Question 25.

Simplify

Answer:

⇒ 

⇒ 

⇒ 

⇒ 

⇒ sec θ – sec θ

⇒ 0

Question 26.

Simplify

Answer:

⇒ 

⇒ 

⇒ 

⇒ 

⇒ sec θ – sec θ

⇒ 0

Question 27.

Simplify

Answer:

⇒ 

⇒ 

⇒ 1 + 1 –(1/2)

⇒ 3/2

Question 28.

Simplify

Answer:

⇒ 

⇒ 

⇒ 1 + 1 –(1/2)

⇒ 3/2

Question 29.

Simplify

cot12° cot38° cot52° cot60° cot78°.

Answer:

⇒ cot 12° cot 38° tan (90–52)° tan (90–78)°

⇒ cot 12° cot 38° tan (38)° tan (12)°

⇒ cot 12° × 1 × tan (12)°

⇒ 1

Question 30.

Simplify

cot12° cot38° cot52° cot60° cot78°.

Answer:

⇒ cot 12° cot 38° tan (90–52)° tan (90–78)°

⇒ cot 12° cot 38° tan (38)° tan (12)°

⇒ cot 12° × 1 × tan (12)°

⇒ 1

Question 31.

Find A if

(i) sin A = cos 30°

(ii) tan49° = cot A

(iii) tan A tan 35° = 1

(iv) sec 35° = cosec A

(v) cosec A cos 43° = 1

(vi) sin 20° tan A sec 70° = √3

Answer:

(i) sin A = sin (90–30)°

⇒ sin A = sin 60°

⇒ A = 60

(ii) cot(90–49) = cot A

⇒ cot 41 = cot A

⇒ A = 41

(iii) tan A = 1/ tan35°

⇒ tan A = cot 35°

⇒ tan A = tan (90–35)

⇒ tan A = tan 55

⇒ A = 55

(iv)cosec(90–35) = cosec A

⇒ cosec 55 = cosec A

⇒ A = 55

(v) cosec A = 1/ cos 43°

⇒ cosec A = sec 43°

⇒ cosec A = cosec (90–43)

⇒ cosec A = cosec 47

⇒ A = 47

(vi) sin 20° tan A cosec(90–70) = √3

⇒ sin 20° tan A cosec 20° = √3

⇒ tan A = √3

⇒ tan A = tan 60°

⇒ A = 60°

Question 32.

Find A if

(i) sin A = cos 30°

(ii) tan49° = cot A

(iii) tan A tan 35° = 1

(iv) sec 35° = cosec A

(v) cosec A cos 43° = 1

(vi) sin 20° tan A sec 70° = √3

Answer:

(i) sin A = sin (90–30)°

⇒ sin A = sin 60°

⇒ A = 60

(ii) cot(90–49) = cot A

⇒ cot 41 = cot A

⇒ A = 41

(iii) tan A = 1/ tan35°

⇒ tan A = cot 35°

⇒ tan A = tan (90–35)

⇒ tan A = tan 55

⇒ A = 55

(iv)cosec(90–35) = cosec A

⇒ cosec 55 = cosec A

⇒ A = 55

(v) cosec A = 1/ cos 43°

⇒ cosec A = sec 43°

⇒ cosec A = cosec (90–43)

⇒ cosec A = cosec 47

⇒ A = 47

(vi) sin 20° tan A cosec(90–70) = √3

⇒ sin 20° tan A cosec 20° = √3

⇒ tan A = √3

⇒ tan A = tan 60°

⇒ A = 60°

Question 33.

Show that

cos 48° – sin 42° = 0

Answer:

Simplifying the left hand side:

⇒ cos 48° – cos(90–42)

⇒ cos 48° – cos(48)°

⇒ 0

Which is equal to right hand side.

Hence proved.

Question 34.

Show that

cos 48° – sin 42° = 0

Answer:

Simplifying the left hand side:

⇒ cos 48° – cos(90–42)

⇒ cos 48° – cos(48)°

⇒ 0

Which is equal to right hand side.

Hence proved.

Question 35.

Show that

cos 20° cos 70° – sin 70° sin 20° = 0

Answer:

Simplifying the left hand side:

⇒ cos 20° cos 70° – cos(90–70)° cos (90–20)°

⇒ cos 20° cos 70° – cos 20° cos 70°

⇒ 0

Which is equal to right hand side.

Hence proved.

Question 36.

Show that

cos 20° cos 70° – sin 70° sin 20° = 0

Answer:

Simplifying the left hand side:

⇒ cos 20° cos 70° – cos(90–70)° cos (90–20)°

⇒ cos 20° cos 70° – cos 20° cos 70°

⇒ 0

Which is equal to right hand side.

Hence proved.

Question 37.

Show that

sin (90° – θ)tan θ = sin θ

Answer:

Simplifying the left hand side:

⇒ cos θ tan θ

⇒ cos θ (sin θ/cos θ)

⇒ sin θ

Which is equal to right hand side.

Hence proved.

Question 38.

Show that

sin (90° – θ)tan θ = sin θ

Answer:

Simplifying the left hand side:

⇒ cos θ tan θ

⇒ cos θ (sin θ/cos θ)

⇒ sin θ

Which is equal to right hand side.

Hence proved.

Question 39.

Show that

Answer:

Simplifying the left hand side:

⇒ 

⇒ 

⇒ 1

Which is equal to right hand side.

Hence proved.

Question 40.

Show that

Answer:

Simplifying the left hand side:

⇒ 

⇒ 

⇒ 1

Which is equal to right hand side.

Hence proved.

---

Exercise 2.3

Question 1.

Find the value of the following.

sin 26^o

Answer:

The relevant part of the sine table is given below.

From the table,

We have sin 26° = 0.43837 ≈ 0.4384

Question 2.

Find the value of the following.

sin 26^o

Answer:

The relevant part of the sine table is given below.

From the table,

We have sin 26° = 0.43837 ≈ 0.4384

Question 3.

Find the value of the following.

cos 72^o

Answer:

The relevant part of the cosine table is given below.

From the table,

We have cos 72° = 0.3090

Question 4.

Find the value of the following.

cos 72^o

Answer:

The relevant part of the cosine table is given below.

From the table,

We have cos 72° = 0.3090

Question 5.

Find the value of the following.

tan35^o

Answer:

The relevant part of the tangent table is given below.

From the table, we have

We have tan 35° = 0.7002

Question 6.

Find the value of the following.

tan35^o

Answer:

The relevant part of the tangent table is given below.

From the table, we have

We have tan 35° = 0.7002

Question 7.

Find the value of the following.

sin 75^o 15’

Answer:

We write sin 75° 15’ = sin 75° 12’ + 3’.

The relevant part of the sine table is given below.

From the table, we have

Sin 75° 12’ = 0.9668

Mean difference for 3’ = 0.0002

We know that mean difference is to be added in the case of sine.

∴ sin75° 15’ = 0.9668 + 0.0002 = 0.9670

Question 8.

Find the value of the following.

sin 75^o 15’

Answer:

We write sin 75° 15’ = sin 75° 12’ + 3’.

The relevant part of the sine table is given below.

From the table, we have

Sin 75° 12’ = 0.9668

Mean difference for 3’ = 0.0002

We know that mean difference is to be added in the case of sine.

∴ sin75° 15’ = 0.9668 + 0.0002 = 0.9670

Question 9.

Find the value of the following.

sin 12° 12’

Answer:

The relevant part of the sine table is given below.

From the table, we have

Sin 12° 12’ = 0.2113

Question 10.

Find the value of the following.

sin 12° 12’

Answer:

The relevant part of the sine table is given below.

From the table, we have

Sin 12° 12’ = 0.2113

Question 11.

Find the value of the following.

cos 12^o 35’

Answer:

We write cos 12° 35’ = cos 12° 30’ + 5’.

The relevant part of the cosine table is given below.

From the table, we have

cos 12° 30’ = 0.9763

Mean difference for 5’ = 0.0003

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 12° 35’ = 0.9763 - 0.0003 = 0.9760

Question 12.

Find the value of the following.

cos 12^o 35’

Answer:

We write cos 12° 35’ = cos 12° 30’ + 5’.

The relevant part of the cosine table is given below.

From the table, we have

cos 12° 30’ = 0.9763

Mean difference for 5’ = 0.0003

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 12° 35’ = 0.9763 - 0.0003 = 0.9760

Question 13.

Find the value of the following.

cos 40^o 20’

Answer:

We write cos 40° 20’ = cos 40° 18’ + 2’.

The relevant part of the cosine table is given below.

From the table, we have

cos 40° 18’ = 0.7627

Mean difference for 2’ = 0.0004

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 40° 20’ = 0.7627 - 0.0004 = 0.7623

Question 14.

Find the value of the following.

cos 40^o 20’

Answer:

We write cos 40° 20’ = cos 40° 18’ + 2’.

The relevant part of the cosine table is given below.

From the table, we have

cos 40° 18’ = 0.7627

Mean difference for 2’ = 0.0004

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 40° 20’ = 0.7627 - 0.0004 = 0.7623

Question 15.

Find the value of the following.

tan 10^o 26’

Answer:

We write tan 10° 26’ = tan 10° 24’ + 2’.

The relevant part of the tangent table is given below.

From the table, we have

tan 10° 24’ = 0.1835

Mean difference for 2’ = 0.0006

We know that mean difference is to be added in the case of tangent.

∴ tan 10° 26’ = 0.1835 + 0.0006 = 0.1841

Question 16.

Find the value of the following.

tan 10^o 26’

Answer:

We write tan 10° 26’ = tan 10° 24’ + 2’.

The relevant part of the tangent table is given below.

From the table, we have

tan 10° 24’ = 0.1835

Mean difference for 2’ = 0.0006

We know that mean difference is to be added in the case of tangent.

∴ tan 10° 26’ = 0.1835 + 0.0006 = 0.1841

Question 17.

Find the value of the following.

cot 20^o

Answer:

The relevant part of the cotangent table is given below.

From the table, we have

cot 20° = 2.7475

Question 18.

Find the value of the following.

cot 20^o

Answer:

The relevant part of the cotangent table is given below.

From the table, we have

cot 20° = 2.7475

Question 19.

Find the value of the following.

cot 40⁰ 20’

Answer:

The relevant part of the cotangent table is given below.

From the table, we have

cot 40° 20’ = 1.17777 ≈ 1.1778

Question 20.

Find the value of the following.

cot 40⁰ 20’

Answer:

The relevant part of the cotangent table is given below.

From the table, we have

cot 40° 20’ = 1.17777 ≈ 1.1778

Question 21.

Find the value of θ, if

(i) sinθ= 0.7009

(ii) cos θ = 0.9664

(iii) tan θ = 0.3679

(iv) cotθ = 0.2334

(v) tanθ = 63.6567

Answer:

(i) From the sine table, we find 0.7009 is corresponding to sin 44° 30’.

⇒ sin 44° 30’ = 0.7009

∴ θ = 44° 30’

(ii) From the cosine table, we find 0.9664 is corresponding to cos 14° 54’.

⇒ cos 14° 54’ = 0.9664

∴ θ = 14° 54’

(iii) From the tangent table, we find 0.3679 is corresponding to tan 20° 12’.

⇒ tan 20° 12’ = 0.3679

∴ θ = 20° 12’

(iv) From the cotangent table, we find 0.2334 is corresponding to cot 76° 30’.

⇒ cot 76° 30’ = 0.2334

∴ θ = 76° 30’

(v) From the tangent table, we find 63.6567 is corresponding to tan 89° 6’.

⇒ tan 89° 6’ = 63.6567

∴ θ = 89° 6’

Question 22.

Find the value of θ, if

(i) sinθ= 0.7009

(ii) cos θ = 0.9664

(iii) tan θ = 0.3679

(iv) cotθ = 0.2334

(v) tanθ = 63.6567

Answer:

(i) From the sine table, we find 0.7009 is corresponding to sin 44° 30’.

⇒ sin 44° 30’ = 0.7009

∴ θ = 44° 30’

(ii) From the cosine table, we find 0.9664 is corresponding to cos 14° 54’.

⇒ cos 14° 54’ = 0.9664

∴ θ = 14° 54’

(iii) From the tangent table, we find 0.3679 is corresponding to tan 20° 12’.

⇒ tan 20° 12’ = 0.3679

∴ θ = 20° 12’

(iv) From the cotangent table, we find 0.2334 is corresponding to cot 76° 30’.

⇒ cot 76° 30’ = 0.2334

∴ θ = 76° 30’

(v) From the tangent table, we find 63.6567 is corresponding to tan 89° 6’.

⇒ tan 89° 6’ = 63.6567

∴ θ = 89° 6’

Question 23.

Simplify, using trigonometric tables

sin 30°30’+cos 40°20’

Answer:

First consider sin 30° 30’,

The relevant part of the sine table is given below.

From the table, we have

Sin 30° 30’ = 0.5075

Then consider cos 40° 20’,

We write cos 40° 20’ = cos 40° 18’ + 2’.

The relevant part of the cosine table is given below.

From the table, we have

cos 40° 18’ = 0.7627

Mean difference for 2’ = 0.0004

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 40° 20’ = 0.7627 - 0.0004 = 0.7623

∴ sin 30° 30’ + cos 40° 20’ = 0.5075 + 0.7623 = 1.2698

Question 24.

Simplify, using trigonometric tables

sin 30°30’+cos 40°20’

Answer:

First consider sin 30° 30’,

The relevant part of the sine table is given below.

From the table, we have

Sin 30° 30’ = 0.5075

Then consider cos 40° 20’,

We write cos 40° 20’ = cos 40° 18’ + 2’.

The relevant part of the cosine table is given below.

From the table, we have

cos 40° 18’ = 0.7627

Mean difference for 2’ = 0.0004

We know that mean difference is to be subtracted in the case of cosine.

∴ cos 40° 20’ = 0.7627 - 0.0004 = 0.7623

∴ sin 30° 30’ + cos 40° 20’ = 0.5075 + 0.7623 = 1.2698

Question 25.

Simplify, using trigonometric tables

tan 45° 27’ + sin 20°

Answer:

First we consider tan 45° 27’.

We write tan 45° 27’ = tan 45° 24’ + 3’.

The relevant part of the tangent table is given below.

From the table, we have

tan 45° 24’ = 1.0141

Mean difference for 3’ = 0.0018

We know that mean difference is to be added in the case of tangent.

∴ tan 45° 27’ = 1.0141 + 0.0018 = 1.0159

Then, consider sin 20°,

The relevant part of the sine table is given below.

From the table, we have

Sin 20° = 0.3420

∴ tan 45° 27’ + sin 20° = 1.0159 + 0.3240 = 1.3399

Question 26.

Simplify, using trigonometric tables

tan 45° 27’ + sin 20°

Answer:

First we consider tan 45° 27’.

We write tan 45° 27’ = tan 45° 24’ + 3’.

The relevant part of the tangent table is given below.

From the table, we have

tan 45° 24’ = 1.0141

Mean difference for 3’ = 0.0018

We know that mean difference is to be added in the case of tangent.

∴ tan 45° 27’ = 1.0141 + 0.0018 = 1.0159

Then, consider sin 20°,

The relevant part of the sine table is given below.

From the table, we have

Sin 20° = 0.3420

∴ tan 45° 27’ + sin 20° = 1.0159 + 0.3240 = 1.3399

Question 27.

Simplify, using trigonometric tables

tan 63°12’ – cos 12°42’

Answer:

First we consider tan 63° 12’,

The relevant part of the tangent table is given below.

From the table, we have

tan 63° 12’ = 1.9797

Then consider cos 12° 42’,

The relevant part of the cosine table is given below.

From the table, we have

cos 12° 42’ = 0.9755

∴ tan 63° 12’ – cos 12° 42’ = 1.9797 – 0.9755 = 1.0042

Question 28.

Simplify, using trigonometric tables

tan 63°12’ – cos 12°42’

Answer:

First we consider tan 63° 12’,

The relevant part of the tangent table is given below.

From the table, we have

tan 63° 12’ = 1.9797

Then consider cos 12° 42’,

The relevant part of the cosine table is given below.

From the table, we have

cos 12° 42’ = 0.9755

∴ tan 63° 12’ – cos 12° 42’ = 1.9797 – 0.9755 = 1.0042

Question 29.

Simplify, using trigonometric tables

sin 50° 26’ + cos 18° + tan 70° 12’

Answer:

First, we consider sin 50° 26’,

We write sin 50° 26’ = sin 50° 24’ + 2’.

The relevant part of the sine table is given below.

From the table, we have

Sin 50° 24’ = 0.7705

Mean difference for 2’ = 0.0004

We know that mean difference is to be added in the case of sine.

∴ sin50° 26’ = 0.7705 + 0.0004 = 0.7709

Now consider cos 18°,

The relevant part of the cosine table is given below.

From the table, we have

cos 18° = 0.9511

Then consider tan 70° 12’,

The relevant part of the tangent table is given below.

From the table, we have

tan 70° 12’ = 2.7776

∴ sin 50° 26’ + cos 18° + tan 70° 12’ = 0.7709 + 0.9511 + 2.7776 = 4.4996

Question 30.

Simplify, using trigonometric tables

sin 50° 26’ + cos 18° + tan 70° 12’

Answer:

First, we consider sin 50° 26’,

We write sin 50° 26’ = sin 50° 24’ + 2’.

The relevant part of the sine table is given below.

From the table, we have

Sin 50° 24’ = 0.7705

Mean difference for 2’ = 0.0004

We know that mean difference is to be added in the case of sine.

∴ sin50° 26’ = 0.7705 + 0.0004 = 0.7709

Now consider cos 18°,

The relevant part of the cosine table is given below.

From the table, we have

cos 18° = 0.9511

Then consider tan 70° 12’,

The relevant part of the tangent table is given below.

From the table, we have

tan 70° 12’ = 2.7776

∴ sin 50° 26’ + cos 18° + tan 70° 12’ = 0.7709 + 0.9511 + 2.7776 = 4.4996

Question 31.

Simplify, using trigonometric tables

tan 72° + cot 30°

Answer:

First we consider tan 72°,

The relevant part of the tangent table is given below.

From the table, we have

tan 72° = 3.0777

Then consider cot 30°,

The relevant part of the cotangent table is given below.

From the table, we have

cot 30° = 1.73205

∴ tan 72° + cot 30° = 3.0777 + 1.73205 = 4.80975 ≈ 4.8098

Question 32.

Simplify, using trigonometric tables

tan 72° + cot 30°

Answer:

First we consider tan 72°,

The relevant part of the tangent table is given below.

From the table, we have

tan 72° = 3.0777

Then consider cot 30°,

The relevant part of the cotangent table is given below.

From the table, we have

cot 30° = 1.73205

∴ tan 72° + cot 30° = 3.0777 + 1.73205 = 4.80975 ≈ 4.8098

Question 33.

Find the area of the right triangle with hypotenuse 20cm and one of the acute angle is 48°

Answer: 

From the above figure,

⇒ sin θ = 

⇒ sin 48° = 

From the sine table, sin 48° = 0.7431.

∴ 0.7431 = 

∴ AB = 0.7431 × 20 = 14.862 cm

Then cos θ = 

⇒ cos 48° = 

From the cosine table, cos 48° = 0.6691

∴ 0.6691 = 

∴ BC = 0.6691 × 20 = 13.382 cm

We know that Area of right angled triangle = 1/2 bh.

∴ Area of ΔABC = 1/2 × 13.382 × 14.862

= 99.441642

∴ Area of the triangle is 99.441 cm².

Question 34.

Find the area of the right triangle with hypotenuse 20cm and one of the acute angle is 48°

Answer: 

From the above figure,

⇒ sin θ = 

⇒ sin 48° = 

From the sine table, sin 48° = 0.7431.

∴ 0.7431 = 

∴ AB = 0.7431 × 20 = 14.862 cm

Then cos θ = 

⇒ cos 48° = 

From the cosine table, cos 48° = 0.6691

∴ 0.6691 = 

∴ BC = 0.6691 × 20 = 13.382 cm

We know that Area of right angled triangle = 1/2 bh.

∴ Area of ΔABC = 1/2 × 13.382 × 14.862

= 99.441642

∴ Area of the triangle is 99.441 cm².

Question 35.

Find the area of the triangle with hypotenuse 8cm and one of the acute angle is 57°

Answer:

From the above figure,

⇒ sin θ = 

⇒ sin 57° = 

From the sine table, sin 57° = 0.8387.

∴ 0.8387 = 

∴ AB = 0.8387 × 8 = 6.7096 cm

Then cos θ = 

⇒ cos 57° = 

From the cosine table, cos 57° = 0.5446

∴ 0.5446 = 

∴ BC = 0.5446 × 8 = 4.3568 cm

We know that Area of right angled triangle = 1/2 bh.

∴ Area of ΔABC = 1/2 × 4.3568 × 6.7096

= 14.616192

∴ Area of the triangle is 14.62 cm² (Approximately).

Question 36.

Find the area of the triangle with hypotenuse 8cm and one of the acute angle is 57°

Answer:

From the above figure,

⇒ sin θ = 

⇒ sin 57° = 

From the sine table, sin 57° = 0.8387.

∴ 0.8387 = 

∴ AB = 0.8387 × 8 = 6.7096 cm

Then cos θ = 

⇒ cos 57° = 

From the cosine table, cos 57° = 0.5446

∴ 0.5446 = 

∴ BC = 0.5446 × 8 = 4.3568 cm

We know that Area of right angled triangle = 1/2 bh.

∴ Area of ΔABC = 1/2 × 4.3568 × 6.7096

= 14.616192

∴ Area of the triangle is 14.62 cm² (Approximately).

Question 37.

Find the area of isosceles triangle with base 16cm and vertical angle 60° 40’

Answer:

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 60° 40’

Then ∠ACD =  = 30° 20’ = ∠BCD

In right angled triangle ACD,

⇒ tan 30° 20’ = 

⇒ tan 30° 20’ = 

From tangent table, tan 30° 18’ = 0.5844 and Mean difference of 2’ = 0.0008.

∴ tan 30° 20’ = 0.5844 + 0.0008 = 0.5852

⇒ CD = 8 ÷ 0.5852 = 13.672 cm

We know that Area of right angled triangle = 1/2 bh.

⇒ Area of ΔACD = 1/2 × 8 × 13.672 = 54.688 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 54.688 cm².

∴ Area of ΔABC = Area of (ΔACD + ΔBCD)

= 54.688 + 54.688

= 109.376 cm²

∴ The area of given isosceles triangle = 109.376 cm².

Question 38.

Find the area of isosceles triangle with base 16cm and vertical angle 60° 40’

Answer:

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 60° 40’

Then ∠ACD =  = 30° 20’ = ∠BCD

In right angled triangle ACD,

⇒ tan 30° 20’ = 

⇒ tan 30° 20’ = 

From tangent table, tan 30° 18’ = 0.5844 and Mean difference of 2’ = 0.0008.

∴ tan 30° 20’ = 0.5844 + 0.0008 = 0.5852

⇒ CD = 8 ÷ 0.5852 = 13.672 cm

We know that Area of right angled triangle = 1/2 bh.

⇒ Area of ΔACD = 1/2 × 8 × 13.672 = 54.688 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 54.688 cm².

∴ Area of ΔABC = Area of (ΔACD + ΔBCD)

= 54.688 + 54.688

= 109.376 cm²

∴ The area of given isosceles triangle = 109.376 cm².

Question 39.

Find the area of isosceles triangle with base 15cm and vertical angle 80°

Answer:

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 80°

Then ∠ACD =  = 40° = ∠BCD

In right angled triangle ACD,

⇒ tan 40° = 

⇒ tan 40° = 

From tangent table, tan 40° = 0.8391.

⇒ CD = 7.5 ÷ 0.8391 = 8.938 cm

We know that Area of right angled triangle = 1/2 bh.

⇒ Area of ΔACD = 1/2 × 7.5 × 8.938 = 33.5175 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 33.5175 cm².

∴ Area of ΔABC = Area of (ΔACD + ΔBCD)

= 33.5175 + 33.5715

= 67.035 cm²

∴ The area of given isosceles triangle = 67.035 cm².

Question 40.

Find the area of isosceles triangle with base 15cm and vertical angle 80°

Answer:

Draw CD perpendicular to AB.

∴ D is the midpoint of AB and ∠ACB = 80°

Then ∠ACD =  = 40° = ∠BCD

In right angled triangle ACD,

⇒ tan 40° = 

⇒ tan 40° = 

From tangent table, tan 40° = 0.8391.

⇒ CD = 7.5 ÷ 0.8391 = 8.938 cm

We know that Area of right angled triangle = 1/2 bh.

⇒ Area of ΔACD = 1/2 × 7.5 × 8.938 = 33.5175 cm²

Since ΔABC is isosceles, area of ΔACD = area of ΔBCD = 33.5175 cm².

∴ Area of ΔABC = Area of (ΔACD + ΔBCD)

= 33.5175 + 33.5715

= 67.035 cm²

∴ The area of given isosceles triangle = 67.035 cm².

Question 41.

A ladder makes an angle 30° with the floor and its lower end is 12m away from the wall. Find the length of the ladder.

Answer:

Let AC be the length of the ladder.

AB is the distance between the foot of the ladder and the wall = 12m

In ΔABC,

⇒ cos 30° = 

⇒ cos 30° = 

From cosine table, cos 30° = 0.8660

⇒ 0.8660 = 

⇒ AC =  = 13.856 m

∴ The length of the ladder is 13.856 m.

Question 42.

A ladder makes an angle 30° with the floor and its lower end is 12m away from the wall. Find the length of the ladder.

Answer:

Let AC be the length of the ladder.

AB is the distance between the foot of the ladder and the wall = 12m

In ΔABC,

⇒ cos 30° = 

⇒ cos 30° = 

From cosine table, cos 30° = 0.8660

⇒ 0.8660 = 

⇒ AC =  = 13.856 m

∴ The length of the ladder is 13.856 m.

Question 43.

Find the angle made by a ladder of length 4m with the ground if its one end is 2m away from the wall and the other end is on the wall.

Answer:

Let θ be the angle made by the ladder with the ground.

AB is the length of the ladder = 4m

AC is the distance between end of ladder and the wall = 2m

In ΔABC,

⇒ cos θ = 

⇒ cos θ =  = 1/2 = 0.5

From cosine table, we find 0.5 is corresponding to cos 60°.

⇒ cos θ = cos 60°

∴ θ = 60°

∴ The angle made by the ladder with the ground is 60°.

Question 44.

Find the angle made by a ladder of length 4m with the ground if its one end is 2m away from the wall and the other end is on the wall.

Answer:

Let θ be the angle made by the ladder with the ground.

AB is the length of the ladder = 4m

AC is the distance between end of ladder and the wall = 2m

In ΔABC,

⇒ cos θ = 

⇒ cos θ =  = 1/2 = 0.5

From cosine table, we find 0.5 is corresponding to cos 60°.

⇒ cos θ = cos 60°

∴ θ = 60°

∴ The angle made by the ladder with the ground is 60°.

Question 45.

Find the length of the chord of a circle of radius 5cm subtending an angle of 108° at the centre.

Answer:

Let AB be the chord of a circle of radius 5 cm with O as centre.

Draw OC perpendicular to AB.

∴ C is the midpoint of AB and ∠AOB = 108°

Then ∠AOC =  = 54°

In right angled triangle OCA,

⇒ sin 54° = 

⇒ sin 54° = 

⇒ AC = sin 54° × 5

= 0.8090 × 5

= 4.045 cm

∴ Length of the chord AB = AC × 2 = 4.045 × 2 = 8.90 cm

Question 46.

Find the length of the chord of a circle of radius 5cm subtending an angle of 108° at the centre.

Answer:

Let AB be the chord of a circle of radius 5 cm with O as centre.

Draw OC perpendicular to AB.

∴ C is the midpoint of AB and ∠AOB = 108°

Then ∠AOC =  = 54°

In right angled triangle OCA,

⇒ sin 54° = 

⇒ sin 54° = 

⇒ AC = sin 54° × 5

= 0.8090 × 5

= 4.045 cm

∴ Length of the chord AB = AC × 2 = 4.045 × 2 = 8.90 cm

Question 47.

Find the length of the side of regular polygon of 12 sides inscribed in a circle of radius 6cm.

Answer:

Let AB be a side of the regular polygon with 12 sides in the circle of radius 6 cm.

If O is a centre of the circle, then ∠AOB =  = 30°

Draw OC perpendicular to AB.

Then ∠AOC =  = 15°

⇒ sin 15° =  = 

⇒ 0.2588 = 

⇒ AC = 0.2588 × 6 = 1.5528

∴ Length of the side AB = 2 × AC = 2 × 1.5528 = 3.1056 cm

Question 48.

Find the length of the side of regular polygon of 12 sides inscribed in a circle of radius 6cm.

Answer:

Let AB be a side of the regular polygon with 12 sides in the circle of radius 6 cm.

If O is a centre of the circle, then ∠AOB =  = 30°

Draw OC perpendicular to AB.

Then ∠AOC =  = 15°

⇒ sin 15° =  = 

⇒ 0.2588 = 

⇒ AC = 0.2588 × 6 = 1.5528

∴ Length of the side AB = 2 × AC = 2 × 1.5528 = 3.1056 cm

Question 49.

Find the radius of the incircle of a regular hexagon of side 24cm.

Answer:

Let AB be the side of the regular hexagon and let O be the centre of the incircle.

Draw OC perpendicular to AB.

If r is the radius of the circle, then OC = r.

We know that total sum of angles of a regular polygon = 360°

So, ∠AOB =  = 60°

∴ ∠AOC =  = 30°

⇒ tan 30° = 

⇒  = 

⇒ r = 12 × 1.732 = 20.784 cm

Hence, radius of incircle is 20.784 cm.

Question 50.

Find the radius of the incircle of a regular hexagon of side 24cm.

Answer:

Let AB be the side of the regular hexagon and let O be the centre of the incircle.

Draw OC perpendicular to AB.

If r is the radius of the circle, then OC = r.

We know that total sum of angles of a regular polygon = 360°

So, ∠AOB =  = 60°

∴ ∠AOC =  = 30°

⇒ tan 30° = 

⇒  = 

⇒ r = 12 × 1.732 = 20.784 cm

Hence, radius of incircle is 20.784 cm.

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Exercise 2.4

Question 1.

The value of sin²60° + cos²60° is equal to
A. sin²45° + cos²45°

B. tan²45° + cot²45°

C. sec²90°

D. 0

Answer:

We know,

∴ 

⇒ .

Comparing the above result with the given options, we find that

A. is the correct option because,

⇒ 

B. is incorrect because,

⇒ 

C. is incorrect because,

 is not defined.

⇒ 

D. is incorrect because, 

Question 2.

The value of sin²60° + cos²60° is equal to
A. sin²45° + cos²45°

B. tan²45° + cot²45°

C. sec²90°

D. 0

Answer:

We know,

∴ 

⇒ .

Comparing the above result with the given options, we find that

A. is the correct option because,

⇒ 

B. is incorrect because,

⇒ 

C. is incorrect because,

 is not defined.

⇒ 

D. is incorrect because, 

Question 3.

If x = , then the value of x is
A. tan 45°

B. tan 30°

C. tan 60°

D. tan 90°

Answer:

Using  in the given expression,

⇒ 

Comparing the above results with the given options, we find that

A. is incorrect because, 

B. is incorrect because,

C. is the correct option because,

D. is incorrect because,

Question 4.

If x = , then the value of x is
A. tan 45°

B. tan 30°

C. tan 60°

D. tan 90°

Answer:

Using  in the given expression,

⇒ 

Comparing the above results with the given options, we find that

A. is incorrect because, 

B. is incorrect because,

C. is the correct option because,

D. is incorrect because,

Question 5.

The value of sec²45° – tan²45° is equal to
A. sin²60° – cos²60°

B. sin² 45° + cos² 60

C.sec² 60° - tan² 60°

D. 0

Answer:

We know, 

∴ 

⇒ .

Comparing the above result with the given options, we find that

A. is incorrect because,

⇒ 

B. is incorrect because,

⇒ 

C. is the correct option because,

⇒ 

D. is incorrect because,

Question 6.

The value of sec²45° – tan²45° is equal to
A. sin²60° – cos²60°

B. sin² 45° + cos² 60

C.sec² 60° - tan² 60°

D. 0

Answer:

We know, 

∴ 

⇒ .

Comparing the above result with the given options, we find that

A. is incorrect because,

⇒ 

B. is incorrect because,

⇒ 

C. is the correct option because,

⇒ 

D. is incorrect because,

Question 7.

The value of 2sin30° cos30° is equal to
A. tan 30°

B. cos 60°

C. sin 60°

D. cot 60°

Answer:

We know, 

∴ 

⇒ 

Comparing the above result with the given options, we find that

A. is incorrect because, 

B. is incorrect because,

C. is the correct option because,

D. is incorrect because,

Question 8.

The value of 2sin30° cos30° is equal to
A. tan 30°

B. cos 60°

C. sin 60°

D. cot 60°

Answer:

We know, 

∴ 

⇒ 

Comparing the above result with the given options, we find that

A. is incorrect because, 

B. is incorrect because,

C. is the correct option because,

D. is incorrect because,

Question 9.

The value of cosec² 60° – 1 is equal to
A. cos² 60°

B. cot² 60°

C. sec² 60°

D. tan² 60°

Answer:

We know, 

∴ 

⇒ 

Comparing the above result with the given options, we find that

A. is incorrect because, 

B. is the correct option because,

C. is incorrect because,

D. is incorrect because,

Question 10.

The value of cosec² 60° – 1 is equal to
A. cos² 60°

B. cot² 60°

C. sec² 60°

D. tan² 60°

Answer:

We know, 

∴ 

⇒ 

Comparing the above result with the given options, we find that

A. is incorrect because, 

B. is the correct option because,

C. is incorrect because,

D. is incorrect because,

Question 11.

cos60° cos30° – sin60° sin30° is equal to
A. cos 90°

B. cosec 90°

C. sin 30° + cos 30°

D. tan 90°

Answer:

We know,  and 

∴ 

⇒ .

Comparing the above result with the given options, we find that

A. is the correct because, 

B. is incorrect because,

C. is incorrect because,

D. is incorrect because,

Question 12.

cos60° cos30° – sin60° sin30° is equal to
A. cos 90°

B. cosec 90°

C. sin 30° + cos 30°

D. tan 90°

Answer:

We know,  and 

∴ 

⇒ .

Comparing the above result with the given options, we find that

A. is the correct because, 

B. is incorrect because,

C. is incorrect because,

D. is incorrect because,

Question 13.

The value of  is
A.0

B. 1

C. tan 27°

D. cot 63°

Answer:

We know, 

∴ 

⇒

Hence option B is correct.

Note: Alternatively, the identity could also have been used.

Question 14.

The value of  is
A.0

B. 1

C. tan 27°

D. cot 63°

Answer:

We know, 

∴ 

⇒

Hence option B is correct.

Note: Alternatively, the identity could also have been used.

Question 15.

If cos x = sin 43°, then the value of x is
A. 57°

B. 43°

C. 47°

D. 90°

Answer:

We know, 

∴ 

So, C is the correct option.

Question 16.

If cos x = sin 43°, then the value of x is
A. 57°

B. 43°

C. 47°

D. 90°

Answer:

We know, 

∴ 

So, C is the correct option.

Question 17.

The value of sec 29° – cosec 61° is
A. 1

B. 0

C. sec 60°

D. cosec 29°

Answer:

We know, 

∴ 

⇒ 

Using the above result, 

So, B is the correct option.

Note: Alternatively, the identity could also have been used.

Question 18.

The value of sec 29° – cosec 61° is
A. 1

B. 0

C. sec 60°

D. cosec 29°

Answer:

We know, 

∴ 

⇒ 

Using the above result, 

So, B is the correct option.

Note: Alternatively, the identity could also have been used.

Question 19.

If 3x cosec 36° = sec 54°, then the value of x is
A. 0

B. 1

C. 

D. 

Answer:

It is given that, 

Using 

∴ 

⇒

And so,  can be rewritten as

i.e 

∴ 

So, C is the correct option.

Note: Alternatively, the identity could also have been used.

Question 20.

If 3x cosec 36° = sec 54°, then the value of x is
A. 0

B. 1

C. 

D. 

Answer:

It is given that, 

Using 

∴ 

⇒

And so,  can be rewritten as

i.e 

∴ 

So, C is the correct option.

Note: Alternatively, the identity could also have been used.

Question 21.

The value of sin60° cos30° + cos60° sin30° is equal to
A. sec 90°

B. tan 90°

C. cos 60°

D. sin 90°

Answer:

We know,  and 

∴ 

⇒ .

Comparing the above result with the given options, we find that

A. is incorrect because, 

B. is incorrect because,

C. is incorrect because,

D. is the correct because,

Question 22.

The value of sin60° cos30° + cos60° sin30° is equal to
A. sec 90°

B. tan 90°

C. cos 60°

D. sin 90°

Answer:

We know,  and 

∴ 

⇒ .

Comparing the above result with the given options, we find that

A. is incorrect because, 

B. is incorrect because,

C. is incorrect because,

D. is the correct because,

Question 23.

If ,then the measure of A is
A. 90°

B. 60°

C. 45°

D. 30°

Answer:

We know,  and 

Given 

So, 

⇒ 

⇒ 

So, B is the correct option.

Question 24.

If ,then the measure of A is
A. 90°

B. 60°

C. 45°

D. 30°

Answer:

We know,  and 

Given 

So, 

⇒ 

⇒ 

So, B is the correct option.

Question 25.

The value of  is
A. 

B. 

C. 0

D. 1

Answer:

We know, 

Using the above result, 

So, D is the correct option.

Note: Alternatively, the identity could also have been used.

Question 26.

The value of  is
A. 

B. 

C. 0

D. 1

Answer:

We know, 

Using the above result, 

So, D is the correct option.

Note: Alternatively, the identity could also have been used.

Question 27.

The value of sin 60° – cos 30° is
A. 0

B. 

C. 

D. 1

Answer:

We know, 

So, 

So, A is the correct option.

Question 28.

The value of sin 60° – cos 30° is
A. 0

B. 

C. 

D. 1

Answer:

We know, 

So, 

So, A is the correct option.

Question 29.

The value of cos²30° – sin²30° is
A. cos 60°

B. sin 60°

C. 0

D. 1

Answer:

We know, 

So, 

⇒ 

Comparing the above result with the given options, we find that

A. is the correct option because, 

B. is incorrect because, 

C. is incorrect because,

D. is incorrect because,

Question 30.

The value of cos²30° – sin²30° is
A. cos 60°

B. sin 60°

C. 0

D. 1

Answer:

We know, 

So, 

⇒ 

Comparing the above result with the given options, we find that

A. is the correct option because, 

B. is incorrect because, 

C. is incorrect because,

D. is incorrect because,