RS Aggarwal Solutions Chapter 8 Trigonometric Identities MCQ Class 10 Maths

RS Aggarwal Solutions Chapter 8 Trigonometric Identities MCQ Class 10 Maths

Chapter Name

RS Aggarwal Chapter 8 Trigonometric Identities

Book Name

RS Aggarwal Mathematics for Class 10

Other Exercises

  • Exercise 8A
  • Exercise 8B
  • Exercise 8C

Related Study

NCERT Solutions for Class 10 Maths

Trigonometric Identities MCQ Solutions

1. (cos56° + cos2 34°)/(sin2 56° + sin2 34°) + 3 tan2 56° tan2 34° = ?

(a) 3.1/2

(b) 4

(c) 6

(d) 5

Solution

(cos2 56° + cos2 34°)/(sin2 56° + sin2 34°) + 3 tan2 56° tan2 34°

= {cos(90° - 34°)}2 + cos2 34°}/{sin(90° - 34°)}2 + sin2 34° + 3{tan(90° - 34°)}2 tan2 34°

= (sin2 34° + cos2 34°)/(cos2 34° + sin2 34°) + 3 cot2 34° tan2 34° 

[∵ cos (90° - θ) = sin θ, sin (90° – θ) = cos θ and tan (90° - θ) = cot θ]

= 1/1 + 3 × 1 [∵ cot θ = 1/tan θ and sin2 θ + cos2 θ = 1]

= 4


2. The value of (sin2 30° cos2 45° + 4 tan2 30° + 1/2 sin2 90° + 1/8 cot2 60°) = ?

(a) 3/8

(b) 5/8

(c) 6

(d) 2

Solution

(d) 2

(sin2 30° cos2 45°) + 4 tan2 30° + 1/2 sin2 90° + 1/8 cot2 60°

= 1/22 × 1/(√2)2 + 4 × 1/(√3)2 + 1/2 × 12 + 1/8 × 1/(√3)2 

[∵ sin 30° = 1/2 and cos 45° = 1/√2 and tan 30° = 1/2 and cot 60° = 1/√3]

= 1/4 × 1/2 + 4 × 1/3 + 1/2 + 1/24

= 1/8 + 4/3 + 1/2 + 1/24

= 1/8 + 4/3 + 1/2 + 1/24

= (3 + 32 + 12 + 1)/24

= 48/24

= 2


3. If cos A + cos2 A = 1 then (sin2 A + sin4 A) = ?

(a) 1/2

(b) 2

(c) 1

(d) 4

Solution

cos2 A + A = 1

⇒ cos A = sin2 A ....(i)

Squaring both sides of (i), we get:

cos2 A = sin4 A ...(ii)

Adding (i) and (ii), we get:

sin2 A + sin4 A = cos A + cos2 A

⇒ sin2 A + sin4 A = 1 [∵ cos A + cosA = 1]


4. If sin θ = √3/2 then (cosec θ + cot θ) = ?

(a) (2 + √3)

(b) 2√3

(c) √2

(d) √3

Solution

(d) √3

Given: sin θ = √3/2 and cosec θ = 2/3

cosec2 θ – cot2 θ = 1

⇒ cot2 θ = cosec2 θ – 1

⇒ cot2 θ = 4/3 – 1 [Given]

⇒ cot θ = 1/√3

∴ cosec θ + cot θ = 2/√3 + 1/√3

= 3/√3

= (√3 × √3)/√3

= √3


5. If cot A = 4/5, prove that (sin A + cos A)/(sin A – cos A) = 9.

Solution

Given: cot A = 4/5

Writing cot A = cos A/sin A and squaring the equation, we get:

cos2 A/sin2 A = 16/25

⇒ 25 cos2 A = 16 sin2 A

⇒ 25 cos2 A = 16 – 16 cos2 A

= cos2 A = 16/41

= 9/1 
= 9 
= RHS

6. If 2x = sec A and 2/x = tan A, prove that (x2 – 1/x2) = 1/4.

Solution

Given:

2x = sec A

⇒ x = sec A/2 ...(i)

and 2/x = tan A

⇒ 1/x = tan A ...(ii)

∴ x + 1/x = sec A/2 + tan A/2 [∵ From (i) and (ii)]

Also, x – 1/x = sec A/2 – tan A/2

∴ (x + 1/x)(x – 1/x) = (sec A/2 + tan A/2)(sec A/2 – tan A/2)

⇒ x2 – 1/x2 = 1/4 (sec2 A – tan2 A)

∴ x2 – 1/x2 = 1/4 × 1 (∵ sec2 A – tan2 A = 1)

= 1/4

Hence proved.


7. If √3 tan θ = 3 sin θ, prove that (sin2 θ – cos2 θ) = 1/3.

Solution

Given: √3 tan θ = 3 sin θ

⇒ √3/cos θ = 3 [∵ tan θ = sin θ/cos θ]

⇒ cos θ = √3/2

⇒ cos2 θ = 3/9

∴ sin2 θ = 1 – 3/9

⇒ sin2 θ = 6/9

∴ LHS = sin2 θ – cos2 θ

= 6/9 - 3/9  [∴ sin2 θ = 6/9, cos2 θ = 3/9]

= 3/9

= 1/3

= RHS

Hence proved.


8. Proved that (sin2 73° + sin2 17°)/(cos2 28° + cos2 62°) = 1.

Solution

(sin2 73 + sin2 17°)/(cos2 28° + cos2 62°) = 1

LHS = (sin2 73° + sin2 17°)/(cos2 28° + cos2 62°)

= [sin (90° - 17°)]2 + sin2 17°)/[cos(90° - 62°)]2 + cos2 62°

= cos2 17° + sin2 17°)/(sin2 62° + cos2 62°)

= 1/1 [∵ sin2 + cos2 θ = 1]

= 1

= RHS


9. If 2 sin 2θ = √3, prove that θ = 30°.

Solution

2 sin (2θ) = √3

⇒ sin (2θ) = √3/2

⇒ sin (2θ) = sin (60°)

⇒ 2θ = 60°

⇒ θ = 60°/2

⇒ θ = 30°


10. Prove that = (cosec A + cot A).

Solution

= 1/sin A + cos A/sin A

= cosec A + cot A = RHS

Hence proved.


11. If cosec θ + cot θ = p, prove that cos θ = (p2 – 1)/(p2 + 1).

Solution

cosec θ + cot θ = p

⇒ 1/sin θ + cos θ/sin θ = p

⇒ (1 + cos θ)/sin θ = p

Squaring both sides, we get:

(1 + cos θ/sin θ)2 = p2

⇒ (1 + cos θ/sin2 θ)2 = p2

⇒ (1 + cos θ)2/(1 – cos2 θ) = p2

⇒ (1 + cos θ)2/(1 + cos θ)(1 – cos θ) = p2

⇒ (1 + cos θ )/(1 – cos θ) = p2

⇒ 1 + cos θ = p2(1 – cos θ)

⇒ 1 + cos θ = p2 – p2 cos θ 

⇒ cos θ (1 + p2) = p2 – 1

⇒ cos θ = (p2 – 1)/(p2 + 1)

Hence proved.


12. Prove that (cosec A – cot A)2 = (1 – cos A)/(1 + cos A).

Solution

(cosec A – cot A)2 = (1 – cos A)/(1 + cos A).

LHS = (cosec A – cot A)2

= (1/sin A – cos A/sin A)2

= (1 – cos A/sin A)2

= (1 – cos A)2/sin2 A

= (1 – cos A)2/(1 – cos2 A)  [∵ sin2 θ + cos2 θ = 1]

= (1 – cos A)(1 – cos A)/(1 – cos A)(1 + cos A)

= (1 – cos A)/(1 + cos A)

= RHS

Hence proved.


13. If 5 cot θ = 3, show that the value of (5 sin θ – 3 cos θ)/(4 sin θ + 3 cos θ) is 16/29.

Solution

Given 5 cot = 3

⇒ 5 cos θ/sin θ = 3 [∵ cot θ = cos θ/sin θ]

⇒ 5 cos θ = 3 sin θ

Squaring both sides, we get:

25 cosθ = 9 sin2 θ

⇒ 25 cos2 θ = 9 – 9 cos2 θ [∵ sinθ + cos2 θ = 1]

⇒ 34 cos2 θ = 9


14. Prove that (sin 32° cos 58°+ cos 32° sin 58°) = 1.

Solution

(sin 32° cos 58° + cos 32° sin 58°) = 1

LHS = sin 32° cos 58° + cos 32° sin 58°

= sin (90° - 58°) cos 58° + cos(90° - 58°) sin 58°

= cos 58° × cos 58° + sin 58° × sin 58° [∵ sin(90° - θ) = cos θ, cos (90° - θ) = sin θ]

= cos2 58° + sin2 58°

= 1 [∵ sin2 θ + cosθ = 1]

= RHS


15. If x = a sin θ + b cos θ and y = a cos θ – b sin θ, prove that x2 + y2 = a+ b2.

Solution

Given: x = a sin θ + b cos θ

Squaring both sides, we get:

x2 = a2 sinθ + 2ab sin θ cos θ + b2 cos2 θ  ...(i)

Also, y = a cos θ – b sin θ

Squaring both sides, we get:

y2 = a2 cos2 θ – 2ab sin θ cos θ + b2 sin2 θ ...(ii)

∴ LHS = x2 + y2

= a2 sin2 + 2absin θ cos θ + b2 cos2 θ + a2 cos2 θ – 2ab sin θ cos θ + b2 sin2 θ 

[using (i) and (ii)]

= a2 (sin2 θ + cos2 θ) + b2(sin2 θ + cos2 θ)

= a2 + b2 [∵ sin2 θ + cos2 θ = 1]

= RHS

Hence proved.


16. Proved that (1 + sin θ)/(1 – sin θ) = (sec θ + tan θ)2.

Solution

(1 + sin θ)/(1 – sin θ) = (sec θ + tan θ)2

LHS = (1 + sin θ)/(1 – sin θ)

Multiplying the numerator and denominator by (1 = sin θ), we get

(1 + sin θ)2/(1 – sin2 θ)

= (1 + 2 sin θ + sin2 θ)/cos2 θ [∵ sin2 θ + cos2 θ = 1]

= secθ + 2 × sin θ/cos θ × sec θ + tan2 θ

= sec2 θ + 2 × tan θ × sec θ + tan2 θ

= (sec θ + tan θ)2

= RHS

Hence proved.


17. Prove that 1/(sec θ – tan θ) – 1/cos θ = 1/cos θ = 1/(sec θ + tan θ).

Solution

1/(sec θ – tan θ) – 1/cos θ = 1/cos θ = 1/(sec θ + tan θ)

LHS = 1/(sec θ – tan θ) – 1/cos θ

= (sec θ + tan θ)/(sec2 θ – tan2 θ) – sec θ [∵ sec2 θ – tan2 θ = 1]

= tan θ

RHS = 1/cos θ – 1/(sec θ + tan θ)

= sec θ – (sec θ – tan θ)/(sec2 θ – tan2 θ) (Multiplying the numerator and denominator by (sec θ – tan θ)

= sec θ + tan θ – sec θ [∵ sec2 θ – tanθ = 1]

= tan θ

∴ LHS = RHS

Hence proved.


18. Prove that (sin A – 2 sin3 A)/(2 cos3 A – cos A) = tan A

Solution

LHS = (sin A – 2 sin3 A)/(2 cos3 A – cos A)

= sin A(1 – 2 sin2 A)/cos A(2 cos2 A – 1)

= tan A{(sin2 A + cos2 A – 2 sin2 A)/(2 cos2 A – sin2 A – cos2 A)} [∵ sin2 A + cos2 A = 1]

= tan A{(cos2 A – sin2 A)/(cos2 A – sinA)}

= tan A

= RHS


19. Prove that tan A/(1 – cot A) + cot A/(1 – tan A) = (1 + tan A + cot A).

Solution

LHS = tan A/(1 – cot A) + cot A/(1 – tan A)

= tan A/(1 – cot A) + cot2 A/(cot A – 1) [∵ tan A = 1/cot A]

= tan A/(1 – cot A) – cot2 A/(1 – cot A)

= (tan A – cot2 A)/(1 – cot A)

= 1/(cot A) – cot2 A/(1 – cot A)

= (1 – cot3 A)/cot A(1 - cot A)

= (1 – cot A)(1 + cot A + cot2 A)/cot A(1 – cot A) [∵ a3 - b3 = (a – b)(a2 + ab + b2)]

= 1/cot A + cot2 A/cot A + cot A/cot A

= 1 + tan A + cot A

= RHS

Hence proved.


20. If sec 5A = cosec (A - 36°) and 5A is an acute angle, show that A = 21°.

Solution

Given: sec 5A = cosec (A - 36°)

⇒ cosec(90° - 5A) = cosec(A - 36°) [∵ cosec (90° - θ) = sec θ]

⇒ 90° - 5A = A - 36°

⇒ 6A = 90° + 36°

⇒ 6A = 126°

⇒ A = 21°


Multiple choice Questions

1. sec 30°/cosec 60° = ?

(a) 2/√3

(b) √3/2

(c) √3

(d) 1

Solution

(d) 1

Sec 30°/cosec 60°

= sec 30°/(sec 90° - 60°) 

= sec 30°/sec 30°

= 1


2. tan 35°/cot 55° + cot 78°/tan 12° = ?

(a) 0

(b) 1

(c) 2

(d) none of these

Solution

(c) 2

We have:

tan 35°/cot 35° + cot 78°/tan 12°

tan 35°/cot(90° - 35°) + cot(90° - 12°)/tan 12°

= tan 35°/tan 35° + tan 12°/tan 12° [∵ cot(90° - θ)]

= 1 + 1

= 2


3. tan 10° tan 15° tan 75° tan 80° = ?

(a) √3

(b) 1/√3

(c) –1

(d) 1

Solution (d) 1

We have,

tan 10° tan 15° tan 75° tan 80°

= tan 10° × tan 15° × tan (90° - 15°) × tan (90° - 10°)

= tan 10° × tan 15° cot 15° × cot 10° [∵ tan(90° - θ) = cot θ]

= 1


4. tan 5° tan 25° tan 30° tan 65° tan 85° = ?

(a) 3

(b) 1/√3

(c) 1

(d) none of these

Solution

(b) 1/√3

We have:

tan 5° tan 25° tan 30°.tan 65° tan 85°

= tan 5° tan 25° tan 30° tan (90° - 25°) tan (90° - 5°)

= tan 5° tan 25° × 1/√3 × cot 25° cot 5° [∵ tan(90° - θ) = cot θ and tan 30° = 1/√3]

= 1/√3


5. cos 1° cos 2° cos 3° ...... cos 180° = ?

(a) -1

(b) 1

(c) 0

(d) 1/2

Solution (c) 0

cos 1° cos2° cos 3° ...... cos 180°

cos 1° cos 2° cos 3° .......cos 90° .......cos (180)°

= 0 [∵ cos 90° = 0]


6. (2 sin2 63° + 1 + 2 sin2 27°)/(3 cos2 17° - 2 + 3 cos2 73°) = ?

(a) 3/2

(b) 2/3

(c) 2

(d) 3

Solution

(d) 3

Given: (2 sin63° + 1 + 2 sin2 27°)/(3 cos2 17° - 2 + 3 cos2 73°)

= [2(sin2 63° + sin2 27°) + 1]/[3(cos2 17° + cos2 73°) – 2]

= 2[sin2 63° + sin2(90° - 63°) + 1]/3[cos2 17° + sin2 17°] – 2)

= 2(sin2 63° + cos2 63°) + 1/3(cos2 17° + sin2 17°) – 2

[∵ sin(90° - θ) = cos θ and cos (90° - θ) = sin θ]

= (2 × 1 + 1)/(3 × 1 – 2)   [∵ sin2 θ + cos2 θ = 1]

= (2 + 1)/(3 – 2)

= 3/1

= 3


7. sin 47° cos 43° + cos 47° sin 43° = ?

(a) sin 4°

(b) cos 4°

(c) 1

(d) 0

Solution

(c) 1

(sin 43° cos 47° + cos 43° sin 47°)

= sin 43° cos (90° - 43°) + cos 43° sin(90° - 43°)

= sin 43° sin 43° + cos 43° sin 43° [∵ cos(90° - θ) = sin θ and sin(90° - θ) = cos θ]

= sin2 43° + cos2 43°

= 1


8. sec 70° sin 20° + cos 20° cosec 70° = ?

(a) 0

(b) 1

(c) -1

(d) 2

Solution

(d) 2

We have:

sec 70° sin 20° + cos 20°cosec 70°

= sin 20°/cos 70° + cos 20°/sin 70°

= sin 20°/cos(90° - 20°) + cos 20°/sin(90° - 20°)

= sin 20°/sin 20° + cos 20°/cos 20° [∵ cos(90° - θ) = sin θ and sin (90° - θ) = cos θ]

= 1 + 1 

= 2

OR

sec 70° sin 20° + cos 20° cosec 70°

= cosec(90° - 70°) sin 20° + cos 20° sec (90° - 70°)

= cosec 20° sin 20° + cos 20° sec 20°

= 1/sin 20° × sin 20° + cos 20° × 1/cos2 0°

= 1 + 1

= 2


9. If sin 3A = cos (A - 10°) and 3A is acute then A = ?

(a) 35°

(b) 25°

(c) 20°

(d) 45°

Solution

(b) 25°

We have:

[sin 3A = cos(A - 10°)]

⇒ cos(90° - 3A) = cos(A - 10°) [∵ sin θ = cos(90° - θ)]

⇒ 90° - 3A = A - 10°

⇒ -4A = - 100

⇒ A = 100/4

⇒ A = 25°


10. If sec 4A = cosec (A - 10°) and 4A is acute then A = ?

(a) 20°

(b) 30°

(c) 30°

(d) 50°

Solution

(a) 20°

We have,

sec 4A = cosec(A - 10°)

⇒ cosec (90° - 4A) = cosec(A - 10°)

Comparing both sides, we get

90° - 4A = A - 10°

⇒ 4A + A = 90° + 10°

⇒ 5A = 100°

⇒ A = 100°/5

∴  A = 20°


11. If A and B are acute angles such that sin A = cos B then (A + B) = ?

(a) 45°

(b) 60°

(c) 90°

(d) 180°

Solution 

(c) 90°


12. If cos (α + β) = 0 then sin (α – β) = ?

(a) sin α

(b) cos β

(c) sin 2α

(d) cos 2β

Solution

(d) cos 2β

We have:

cos (α + β) = 0

⇒ cos(α + β) = cos 90°

⇒ α + β = 90°

⇒ α = 90° - β ...(i)

Now, sin (α – β)

= sin [(90° - β) – β] [Using (i)]

= sin (90° - 2β)

= cos 2β  [∵ sin(90° - θ) = cos θ]


14. sec2 10° - cot2 80°

(a) 1

(b) 0

(c) 3/2

(d) 1/2

Solution

(a) 1

We have (sin 79° cos 11° + cos 79° sin 11°)

= sin 79° cos(90° - 79°) + cos 79° sin(90° - 79°)

= sin 79° sin 79° + cos 79° cos 79° [∵ cos (90° - θ) = sin θ and sin(90° – θ) = cos θ]

= sin2 79° + cos2 79°

= 1


15. cosec2 57° - tan2 33° = ?

(a) 0

(b) 1

(c) – 1

(d) 2

Solution

We have:

(cosec2 57° - tan33°)

= [cosec2(90° - 33°) – tan2 33°]

= (sec33° - tan2 33°) [∵ cosec(90° - θ) = sec θ]

= 1 [∵ sec2 θ – tan2 θ = 1]


16. 2 tan2 30° sec52° sin2 38°)/(cosec2 70° - tan2 20°) = ?

(a) 2

(b) 1/2

(c) 2/3

(d) 3/2

Solution

We have:

[(2 tan2 30° sec2 52° sin2 38°)/(cosec2 70° - tan2 20°)]

= [2 × (1/√3)2 sec2 52°[sin2(90° - 52°)}/{cosec2 (90° - 20°)} – tan2 20°]

= [2/3 × (sec2 52°.cos2 52°/sec2 20° - tan2 20°)] [∵ sin (90° - θ) = cos θ and cosec(90° - θ) = sec θ]

= 2/3 × 1/1 [∵ sec2 θ – tan2 θ = 1]

= 2/3


17. {(sin2 22° + sin2 68°)/(cos2 22° + cos2 68°) + sin2 63° + cos 63° sin 27°} = ?

(a) 0

(b) 1

(c) 2

(d) 3

Solution

We have:

[(sin2 22° + sin2 68°/cos2 22° + cos2 68°) + sin2 63° + cos 63° sin 27°]

= {sin2 22° + sin2(90° - 22°)/cos2(90° - 68°) + cos2 68°} + sin2 63° + cos 63° {sin(90° - 63°)}]

= [sin2 22° + cos2 22°/sin2 68° + cos2 68° + sin2 63° + cos 63° cos 63°] [∵ sin(90° - θ)]

= cos θ and cos (90° - θ) = sin θ]

= [1/1 + sin2 63° + cos2 63°] [∵ sin2 θ + cos2 θ = 1]

= 1 + 1

= 2


18. [cot(90° - θ).sin(90° - θ)]/sin θ + cot 40°/tan 50° - (cos2 20° + cos2 70°) = ?

(a) 0

(b) 1

(c) – 1

(d) none of these

Solution

We have:

[cot(90° - θ). sin(90° - θ)/sin θ + cot 40°/tan 50° - (cos2 20° + cos2 70°)]

= tan θ. cos θ/cos θ + cot (90° - 50°)/tan 50° - {cos2 (90° - 70°) + cos70°}] [∵ cot (90° - θ) = tan θ and sin(90° - θ) = cos θ]

= [sin θ/sin θ + tan 50°/tan 50° - (sin2 70° + cos70°)] [∵ cos(90° - θ) = sin θ]

= (sin θ/sin θ + 1 – 1)

= 1 + 1 – 1

= 1


19. (cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55° = ?

(a) √3

(b) 1/3

(c) 1/√3

(d) 2/√3

Solution

(c) 1/√3

We have:

[(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)]

= cos 38° cosec(90° - 38°)/[tan 18° tan 35° × √3 × tan(90° - 18°)tan(90° - 35°)] [∵ cos(90° - θ) = sec θ and tan (90° - θ) = cot θ]

= [cos 38° sec 38°]/[tan 18° tan 35° × √3 × cot 18° cot 35°]

= [1/sec 38° × sec 38°]/[1/(cot 18° cot 35°) × √3 cot 18° cot 35°]

= 1/√3


20. If 2 sin 2θ = √3 then θ = ?

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Solution

(a) 30°

2 sin 2θ = √3

⇒ sin 2θ = √3/2 = sin 60°

⇒ sin 2θ = sin 60°

⇒ 2θ = 60°

⇒ θ = 30°


21. If 2 cos 3θ = 1 then θ = ?

(a) 10°

(b) 15°

(c) 20°

(d) 30°

Solution

2 cos 3θ = 1

 ⇒ cos 3θ = 1/2

⇒ cos 3θ = cos 60° [∵ cos 60° = 1/2]

⇒ 3θ = 60°

⇒ θ = 60°/3 = 20°


22. If √3 tan 2θ – 3 = 0 then θ = ?

(a) 15°

(b) 30°

(c) 45°

(d) 60°

Solution

√3 tan 2θ – 3 = 0

⇒ √3 tan 2θ = 3

⇒ tan 2θ = 3/√3 

⇒ tan 2θ = √3 [∵ tan 60° = √3]

⇒ tan 2θ = tan 60°

⇒ 2θ = 60°

⇒ θ = 30°


23. If tan x = 3 cot x then x = ?

(a) 45°

(b) 60°

(c) 30°

(d) 15°

Solution

(b) 60°

tan x = 3 cot x

⇒ tan x/cot x = 3

⇒ tan2 x = 3 [∵ cot x = 1/tan x]

⇒ tan x = √3 = tan 60°

⇒ x = 60°


24. If x tan 45° cos 60° = sin 60° then x = ?

(a) 1

(b) 1/2

(c) 1/√2

(d) √3

Solution

(a) 1

x tan 45° cos 60° = sin 60° cot 60°

⇒ x(1)(1/2) = (√3/2)(1/√3)

⇒ x(1/2) = (1/2)

⇒ x = 1


25. If tan2 45° - cos2 30° = x sin 45° cos 45° then x = ?

(a) 2

(b) -2

(c) 1/2

(d) -1/2

Solution

(c) 1/2

(tan2 45° - cos2 30°) = x sin 45° cos 45°

⇒ x = (tan2 45° - cos2 30°)/(sin 45° cos 45°)

= [(1)2 – (√3/2)2]/(1/√2 × 1/√2)

= (1 – 3/4)/(1/2)

= (1/4)/(1/2)

= 1/4 × 2

= 1/2


26. sec60° - 1 = ?

(a) 2

(b) 3

(c) 4

(d) 0

Solution

(b) 3

sec2 60° - 1

= (2)2 – 1

= 4 – 1

= 3


27. (cos 0° + sin 30° + sin 45°)(sin 90° + cos 60° - cos 45°) = ?

(a) 5/6

(b) 5/8

(c) 3/5

(d) 7/4

Solution

(a) 7/4

(cos 0° + sin 30° + sin 45°)(sin 90° + cos 60° - cos 45°)

= (1 + 1/2 + 1/√2)(1 + 1/2 – 1/√2)

= (3/2 + 1/√2)(3/2 – 1/√2)

= [(3/2)2 – (1/√2)2]

= (9/4) – (1/2)

= (9 – 2)/4

= 7/4


28. sin2 30° + 4 cot2 45° - sec2 60° = ?

(a) 0

(b) 1/4

(c) 4

(b) 1

Solution

(b) 1/4

(sin2 30° + 4 cot2 45° - sec2 60°)

= [(1/2)2 + 4 × (1)2 – (2)2]

= (1/4 + 4 – 4)

= 1/4


29. 3 cos2 60° + 2 cot2 30° - 5 sin2 45° = ?

(a) 13/6

(b) 17/4

(c) 1

(d) 4

Solution

(b) 17/4

(3 cos2 60° + 2 cot2 30° - 5 sin2 45°)

= [3 × (1/2)2 + 2 × (√3)2 – 5 × (1/√2)2]

= [3/4 + 6 – 5/2]

= (3 + 24 – 10)/4

= 17/4


30. cos2 30° cos2 45° + 4 sec2 60° + 1/2 cos2 90° - 2 tan2 60° = ?

(a) 73/8

(b) 75/8

(c) 81/8

(d) 83/8

Solution

(d) 83/8

(cos2 30° cos2 45° + 4 sec2 60° + 1/2 cos2 90° - 2 tan2 60°)

= [(√3/2)2 × (1/√2)2 + 4 × (2)2 + 1/2 × (0)– 2 × (√3)2]

= [(3/4 × 1/2) + 16 – 6]

= [3/8 + 10]

= (3 + 80)/8

= 83/8


31. If cosec θ = √10 then sec θ = ?

(a) 

(b) 

(c) 
(d) 

Solution

(b) 

Let us first draw a right ∆ABC right angled at B and ∠A = θ


32. If tan θ = 8/15 then cosec θ = ?

(a) 17/8

(b) 8/17

(c) 17/15

(d) 15/17

Solution

(a) 17/8

Let us first draw a right ∆ABC right angled at B and ∠A = θ.

Give: tan θ = 8/5, but tan θ = BC/AB

So, BC/AB = 8/15

Thus, BC = 8k and AB = 15k

Using Pythagoras theorem in triangle ABC, we have:

AC2 = AB2 + BC2

⇒ AC2 = (15k)2 + (8k)2

⇒ AC2 = 289k2

⇒ AC = 17k

∴ cosec θ = AC/BC = 17k/8k

= 17/8


33. If sin θ = b/a then cos θ = ?

(a) 

(b) 

(c) 

(d) b/a

Solution

(b) 

Let us draw a right ∆ABC right angled at B and A = θ.

Given: sin θ = a/b, but sin θ = BC/AC

So, BC/AC = a/b

Thus, BC = ak and AC = bk

Using Pythagoras theorem in triangle ABC, we have:

AC2 = AB2 + BC2

 AB= AC2 – BC2

 AB2 = (bk)2 – (ak)2

 AB= (b2 – a2)k2



34. If tan θ = √3 then sec θ= ? 

(a) 2/√3

(b) √3/2

(c) 1/2

(d) 2

Solution

(d) 2

Let us first draw a right ∆ABC right angled at B and ∠A = θ.

Give: tan θ = √3

But tan θ = BC/AB

so, BC/AB = √3/1

Thus, BC = √3k and AB = k

Using Pythagoras theorem, we get:

AC2 = AB2 + BC2

⇒ AC2 = (√3k)2 + (k)2

⇒ AC2 = 4k2

⇒ AC = 2k

∴ sec θ = AC/AB = 2k/k = 2/1


35. If sec θ = 25/7 then sin θ = ?

(a) 7/24

(b) 24/7

(c) 24/25

(d) none of these

Solution

(c) 24/25

Let us first draw a right ∆ABC right angled at B and ∠A = θ.

Given sec θ = 25/7

But cos θ = 1/sec θ = AB/AC = 7/25

Thus, AC = 25k and AB = 7k

Using Pythagoras theorem, we get:

AC2 = AB+ BC2

⇒ BC2 = AC2 – AB2

⇒ BC2 = (25k)2 – (7k)2

⇒ BC2 = 576k2

⇒ BC = 24k

∴ sin θ = BC/AC = 24k/25k

= 24/25


36. If sin θ = 1/2 then cot θ = ?

(a) 1/√3

(b) √3

(c) √3/2

(d) 1

Solution

(b) √3

Given: sin θ = 1/2, but sin θ = BC/AC

So, BC/AC = 1/2

Thus, BC = k and AC = 2k

Using Pythagoras theorem in triangle ABC, we have:

AC2 = AB+ BC2

AB2 = AC2 – BC2

AB2 = (2k)2 – (k2)

AB2 = 3k2

AB = √3k

So, tan θ = BC/AB = k/√3k = 1/√3

∴ cot θ = 1/ tan θ = √3


37. If cos θ = 4/5 then tan θ = ?

(a) 3/4

(b) 4/3

(c) 3/5

(d) 5/3

Solution

(a) 3/4

Since cos θ = 4/5 but cos θ = AB/AC

So, AB/AC = 4/5

Thus, AB = 4k and AC = 5k

Using Pythagoras theorem in triangle ABC, we have:

AC2 = AB2 + BC2

⇒ BC2 = (5k)2 – (4k)2

= BC2 = 9k2

⇒ BC = 3k

∴ tan θ = BC/AB = 3/4


38. If 3x = cosec θ and 3/x = cot θ then (x2 – 1/x2) = ?

(a) 1/27

(b) 1/81

(c) 1/3

(d) 1/9

Solution

(c) 1/3

Given: 3x = cosec θ and 3/x = cot θ

Also, we can deduce that x = cosec θ/3 and 1/x = cot θ/3

So, substituting the values of x and 1/x in the given expression, we get:

3(x2 – 1/x2) = 3 (cosec θ/3)2 – (cot θ/3)2)

= 3 (cosec2 θ)/9 – (cot2 θ/9))

3/9 (cosec2 θ – cotθ)

= 1/3  [By using the identity: (cosec2 θ – cot2 θ = 1)]


39. If 2x = sec A and 2/x = tan A then 2(x2 – 1/x2) = ?

(a) 1/2

(b) 1/4

(c) 1/8

(d) 1/16

Solution

(a) 1/2

Given: 2x = sec A and 2/x = tan A

Also, we can deduce that x = sec A/2 and 1/x = tan A/2

So, substituting the values of x and 1/x in the given expression, we get:

2(x2 – 1/x2) = 2(sec A/2)2 – (tan A/2)2

= 2((sec2 A/4) – (tanA/4))

 = 2/4 (sec2 A – tan2 A)

= 1/2 [By using the identity: (sec2 θ – tan2 θ = 1)]


40. If tan θ = 4/3 then (sin θ + cos θ) = ?

(a) 7/3

(b) 7/4

(c) 7/5

(d) 5/7

Solution

(c) 7/5

Let us first draw a right ∆ABC right angled at B and ∠A = θ.

tan θ = 4/3 = BC/AB

So, AB = 3k and BC = 4k

Using Pythagoras theorem, we get:

AC2 = AB2 + BC2

⇒ AC2 = (3k)2 + (4k)2

⇒ AC2 = 25k2

⇒ AC = 5k

Thus, sin θ = BC/AC = 4/5

Ans cos θ = AB/AC = 3/5

∴ (sin θ + cos θ) = (4/5 + 3/5) = 7/5


41. If (tan θ + cot θ) = 5 then (tan2 θ + cot2 θ) = ?

(a) 27

(b) 25

(c) 24

(d) 23

Solution

(d) 23

We have (tan θ + cot θ) = 5

Squaring both sides, we get:

(tan θ + cot θ)2 = 52

⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 25

⇒ tanθ + cot2 θ + 2 = 25  [∵ tan θ = 1/cot θ]

⇒ tan2 θ + cot2 θ = 25 – 2

= 23


42. If (cos θ + sec θ) = 5/2 then (cos2 θ + sec2 θ) = ?

(a) 21/4

(b) 17/4

(c) 29/4

(d) 33/4

Solution

(b) 17/4

Solution

We have (cos θ + sec θ) = 5/2

Squaring both sides, we get:

(cos θ + sec θ)= (5/2)2

⇒ cos2 θ + sec2 θ + 2θ = 25/4

⇒ cosθ + sec2 θ + 2 = 25/4 [∵ sec θ = 1/cos θ]

⇒ cos2 θ + sec2 θ = 25/4 – 2 = 17/4


43. If tan θ = 1/7 then (cosec2 θ + sec2 θ)/(cosec2 θ + sec2 θ) = ?

(a) -2/3

(b) -3/4

(c) 2/3

(d) 3/4

Solution

(d) 3/4

(cosec2 θ – sec2 θ)/(cosec2 θ + sec2 θ)

= sinθ(1/sin2 θ – 1/cos2 θ)/sin2 θ(1/ sin2 θ + 1/cos2 θ) [Multiplying the numerator and denominator by sin2 θ]

= (1 – tan2 θ)/(1 + tan2 θ)

= (1 – 1/7)/(1 + 1/7)

= 6/8

= 3/4


44. If tan θ = 4 then |(7 sin θ - 3 cos θ)/(7 sin θ + 3 cos θ ) = ? 

(a) 1/7

(b) 5/7

(c) 3/7

(d) 5/14

Solution

(a) 1/7

7 tan θ = 4

Now dividing the numerator and denominator of the given expression by cos θ,

We get:

1/cos θ(7 sin θ - 3 cos θ )/1/cos θ(7 sin θ + 3 cos θ)

= (7 tan θ - 3)/(7 tan θ + 3)

= (4 – 3)/(4 + 3)  [∵ tan θ = 4]

= 1/7


45. If 3 cot θ = 4 then (5 sin θ + 3 cos θ)/(5 sin θ + 3 cos θ )/(5 sin θ – 3 cos θ) = ?

(a) 1/3

(b) 3

(c) 1/9

(d) 9

Solution

(d) 9

We have (5 sin θ + 3 cos θ)/(5 sin θ – 3 cos θ)

Dividing the numerator and denominator of the given expression by sin θ, we get:

1/sin θ(5 sin θ + 3 cos θ)/1/sin θ(5 sin θ – 3 cos θ)

= (5 + 3 cot θ)/(5 – 3 cot θ)

= (5 + 4)/(5 – 4) = 9 [∵ 3 cot θ = 4]


46. If tan θ = a/b then (a sin θ – b cos θ)/(a sin θ + b cos θ) = ?

(a) (a2 + b2)/(a2 – b2)

(b) (a2 – b2)/(a2 + b2)

(c) a2/(a2 + b2)

(d) a2/(a2 + b2)

Solution

(b) (a– b2)/(a2 + b2)

We have tan θ = a/b

Now, dividing the numerator and denominator of the given expression by cos θ we get:

(a sin θ – b cos θ)/(a sin θ + b cos θ)

= [1/cos θ(a sin θ – b cos θ)]/[1/cos θ(a sin θ + b cos θ)]

= (a tan θ – b)/(a tan θ + b)

= (a2/b – b)/(a2/b + b)

= (a2 – b2)/(a2 + b2)


47. If sin A + sin2 A = 1 then cos2 A + cos4 A = ?

(a) 1/2

(b) 1

(c) 2

(d) 3

Solution

(b) 1

sin A + sin2 A = 1

⇒ sin A = 1 – sin2 A

⇒ sin A = cos2 A (∵ 1 – sin2 A)

⇒ sin2 A = cos4 A (Squaring both sides)

⇒ 1 – cos2 A = cos4 A

⇒ cos4 A + cos2 A = 1


48. If cos A + cos2 A = 1 then sin2 A + sin4 A = ?

(a) 1

(b) 2

(c) 4

(d) 3

Solution

(a) 1

cos A + cos2 A = 1

⇒ cos A = 1 – cos2 A

⇒ cos A = sin2 A (∵ 1 – cos2 A = sin2)

⇒ cos2 A = sin4 A (Squaring both sides)

⇒ 1 – sin2 A = sin4 A

⇒ sin4 A + sin2 A = 1


49. 

(a) sec A + tan A

(b) sec A – tan A

(c) sec A tan A

(d) none of these

Solution

= (1 – sin A)/cos A

= 1/cos A – sin A/cos A

= sec A – tan A


50. 

(a) cos sec A – cot A

(b) cos sec A – cot A

(c) cosec A cot A

(d) none of these

Solution

(b) cos sec A + cot A

= (1 – cos A)/sin A

= 1/sin A – cos A/sin A

= cosec A – cot A


51. If tan θ = a/b, then (cos θ + sin θ)/(cos θ – sin θ) = ?

(a) (a + b)/(a – b)

(b) (a + b)/(a - b)

(c) (b + a)/(b – a)

(d) (b – a) /(b + a)

Solution

(c) (b + a)/(b – a)

Given: tan θ = a/b

Now, (cos θ + sin θ)/(cos θ – sin θ)

= (1 + tan θ)/(1 – tan θ) [Dividing the numerator and denominator by cos θ ]

= (1 + a/b)/(1 – a/b)

= ((b + a)/b))/((b – a)/b)

= (b + a)/(b – a)


52. (cosec θ – cot θ)2 = ?

(a) (1 + cos θ)/(1 – cos θ)

(b) (1 – cos θ)/(1 + cos θ)

(c) (1 + sin θ)/(1 – sin θ)

(d) none of these

Solution

(b) (1 – cos θ)/(1 + cos θ)

(cosec θ – cot θ)2

= (1/sin θ - cos θ/sin θ)2

= ((1 – cos θ)/sin θ)2

= (1 – cos θ)2/(1 – cos2 θ)

= (1 – cos θ)2/(1 + cos θ)(1 – cos θ)

= (1 – cos θ)/(1 + cos θ)


53. (sec A + tan A)(1 – sin A) = ?

(a) sin A

(b) cos A

(c) sec A

(d) cosec A

Solution

(b) cos A

(sec A + tan A)(1 – sin A)

= (1/cos A + sin A/cos A)(1 – sin A)

= (1 – sin A)/cos A(1 – sin A)

= (1 – sin2 A)/cos A

= cosA/cos A

= cos A

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