Explore topic-wise InterviewSolutions in .

Correct option is: C) Both A and B are True

L.H.S. = 2 sin 45° cos 45°

= 2 × (1/√2) × (1/√2)

= (2 × 1/2)

= 1

R.H.S. = sin 90° = 1

L.H.S. = R.H.S.

sin² θ + cos² θ = 1

i. lf θ = 0°, 

LH.S. = sin² θ + cos² θ 

= sin² 0° + cos² 0° 

= 0 + 1 … [∵ sin 0° = 0, cos 0° = 1] 

= R.H.S. 

∴ sin² θ + cos² θ = 1 

ii. If θ = 90°, 

L.H.S.= sin² θ + cos² θ = sin 90° + cos 90° 

= 1 + 0 … [ ∵ sin 90° = 1, cos 90° = 0] 

= 1 = R.H.S. 

∴ sin² θ + cos² θ = 1

L.H.S. = 2 sin 30° cos 30°

= 2 × (1/2) × (√3/2)

= √3/2

R.H.S.

sin 60° = √3/2

L.H.S. = R.H.S.

L.H.S. = cos 60° cos 30° + sin 60° sin 30°

= (1/2) × (√3/2) + (√3/2)(1/2)

= (√3/4) + (√3/4)

= √3/2

R.H.S.

cos 30° = √3/2

L.H.S. = R.H.S.

We know that,

sin 60° = √3/2 = cos 30°

and sin 30° = 1/2 = cos 60°

Now,

sin 60° cos 30° + cos 60° sin 30° 

= (√3/2)( √3/2) + (1/2)(1/2)

= (3/4) + (1/4)

= 4/4

= 1

L.H.S. = sin 60° cos 30° — cos 60° sin 30°

= (√3/2) × (√3/2) – (1/2)(1/2)

= (3/4) – (1/4)

= 2/4

= 1/2

R.H.S.:

sin 30° = 1/2

LHS = RHS

We know that,

cos 60° = 1/2 = sin 30°

and cos 30° = √3/2 = sin 60°

Now,

cos 60° cos 30° — sin 60° sin 30° 

= (1/2) × (√3/2) – (√3/2) × (1/2)

= (√3/4) – (√3/4)

= 0

LHS = cos54° cos36° − sin54° sin36°

= cos54° cos36° − sin(90° - 36°) sin(90° - 54°)

= cos54° cos36° – cos36°cos54°

= 0

= RHS

cos (A + B) = cos A cos B — sin A sin B

If A = 60° and B = 30°

Verify: cos (90°) = cos 60° cos 30° — sin 60° sin 30°

R.H.S. = cos 60° cos 30° – sin 60° sin 30°

= (1/2) × (√3/2) – (√3/2)(1/2)

= (√3/4) – (√3/4)

= 0

= cos 90°

= L.H.S.

sin (A – B) = sin A cos B – cos A sin B

If A = 60° and B = 30°, then

LHS :

= sin(60°- 30°)

= sin 30°

= 1/2

R.H.S. = sin 60° cos 30° – cos 60° sin 30°

= (√3/2) × (√3/2) – (1/2) (1/2)

= (3/4) – (1/4)

= 2/4

= 1/2

L.H.S. = R.H.S.

Correct option is: A) \(\frac{64}{49}\)

We have tan\(\theta\) = \(\frac{7}{8}\) 

Now, \(\frac{(1+sin \theta)(1-sin \theta)}{1+cos \theta)(1-cos \theta)}\) = \(\frac {1-sin^2\theta}{1-cos^2\theta} = \frac {cos^2\theta}{sin^2\theta} \) (\(\because\)  \(sin^2\theta + cos^2\theta = 1\))

= \((\frac {cos\theta}{sin\theta})^2 = cot^2\theta\) 

= \(\frac {1}{tan^2 \theta} = \frac {1}{(\frac 78)^2} = \frac {8^2}{7^2}\)

= \(\frac {64}{49}\)

Correct option is: A) \(\frac{64}{49}\)

Sin 2 A = 2 sin A 

[∵ 2A = 2 sin A. Cos A] 

⇒ 2 sin A . cos A = 2 sin A 

⇒ cos A = 1 = cos 0° 

⇒ A = 0°

cos (A – B) = cos A cos B + sin A sin B

If A = 60° and B = 30°, then

Verify: cos (30°) = cos 60° cos 30° + sin 60° sin 30°

R.H.S. = cos 60° cos 30° + sin 60° sin 30°

= (1/2) × (√3/2) + (√3/2)(1/2)

= (√3/4) + (√3/4)

= √3/2

= cos 30°

= L.H.S.

Correct option is: A) sin 2A

sin (A + B) cos (A – B) + sin (A – B) cos (A + B)

= sin (A + B) + (A-B))

(\(\because\) sin (x + y) = sin x cos y + cos x sin y, here x = A + B and y = A-B)

= sin (2A)

Correct option is: A) sin 2A

A = 60° and B = 30°

Now, A + B = 60° + 30° = 90° 

Also, A – B = 60° – 30° = 30°

(i) sin (A + B) = sin 90° = 1 

sin A cos B + cos A sin B = sin 60° cos 30° + cos 60° sin 30°

= \((\frac{\sqrt{3}}2\times\frac{\sqrt{3}}2+\frac{1}2\times\frac{1}2)\) = \((\frac{3}4+\frac{1}4)\) = 1

∴ sin (A + B) = sin A cos B + cos A sin B 

(ii) cos (A + B) = cos 90° = 0 

cos A cos B - sin A sin B = cos 60° cos 30° - sin 60° sin 30°

= \((\frac{{1}}2\times\frac{\sqrt{3}}2-\frac{{\sqrt3}}2\times\frac{1}2)\) = \((\frac{\sqrt3}4+\frac{\sqrt3}4)\) = 0

∴ cos (A + B) = cos A cos B - sin A sin B

L.H.S. = cos 2A – cos2B + cos2C 

= -2 sin(A + B) . sin(A – B) + 1 – 2sin²C 

= 1 – 2 sinC sin (A – B) – 2 sin2C [∵ sin(A+B)-sinc] 

= 1 – 2 sinC [sin(A – B) + sinC] 

= 1 – 2 sinC [sin(A + B) + sin(A – B)] 

= 1 – 2 sinC [2sinA . cosB] 

= 1 – 4 sinA cosB sinC = R.H.S.

L.H.S. = cos2A + cos2B + cos2C 

= 2cos(A + B) · cos(A – B) + 2cos2C – 1 

= -2cosC . cos(A – B) + 2cos2C – 1 

= -2cosC[cos (A – B) – cos C] – 1 

= -2cosC{cos (A – B) + cos(A + B)} – 1 

= -2 cosC (2cosA · cosB) – 1 

= – 1 – 4 cos A cosB cosC = R.H.S.

LHS = sin35° sin55° − cos35° cos55°

= sin(90° – 55°) sin(90° – 35°) − cos35° cos55°

= cos55° cos35° – cos35° cos55°

= 0

= RHS

sin² 48° + sin² 42° = 1

LHS = sin² 48° + sin² 42°

= sin² 48° + sin² (90 – 48)°

= sin² 48° + cos² 48°

= 1

= RHS

Correct option is: C) depression

In the given figure \(\theta\) is called angle of depression.

Correct option is: C) depression

Correct option is: B) 60 m

Correct option is: B) Kiran

Correct option is: D) None of these

(A) A = \(30^\circ\) and B = \(60^\circ\)

= A + B = \(30^\circ\) + \(60^\circ\) = \(90^\circ\)

Now, sin A + sin B = sin \(30^\circ\) + sin \(60^\circ\)

= \(\frac 12\) + \(\frac {\sqrt3}{2}\)

= \(\frac {1+\sqrt3}{2} >1\)

But sin (A +B) = sin \(90^\circ\) = 1

Hence, sin A + sin B \(\neq\) sin (A+B)

(B) A = \(30^\circ\), B = \(45^\circ\)

\(\therefore\) A + B = \(30^\circ\) +  \(45^\circ\) = \(75^\circ\)

Now, sin A + sin B = sin \(30^\circ\) + sin \(45^\circ\)

= \(\frac 12\) + \(\frac 1{\sqrt2}\) = \(\frac {1+\sqrt2}2 > 1\)

But sin (A+B) = sin \(75^\circ\) < 1 (\(\because\) -1 \(\leq\) sin \(\theta\) \(\leq\)1)

Hence, sin (A+B) \(\neq\) sin A + sin B

(C) A = \(45^\circ\), B = \(60^\circ\)

\(\therefore\) A + B = \(45^\circ\)+\(60^\circ\) = \(105^\circ\)

Nw, sin A + sin B = sin \(45^\circ\) + sin \(60^\circ\)

= \(\frac 1{\sqrt2}\) + \(\frac {\sqrt3}{2}\) = \(\frac {\sqrt2+\sqrt3}{2}\) >1

But sin (A+B) = sin \(105^\circ\) < 1 ( \(\because\) -1 \(\leq\) sin \(\theta\) \(\leq\) 1)

Hence, sin A + sin B \(\neq\) sin (A+B)

Correct option is: D) None of these

Correct option is: B) Ravi

\(\because\) -1 \(\leq\) cos \(\theta\) \(\leq\) 1

since, 2 > 1

\(\therefore\) cos A never be equal to 2.

Hence, Ramu is saying wrong.

But cos A = 1 (Possible)

Hence, Ravi is saying truth.

We have to agree with Ravi.

Correct option is: B) Ravi

L.H.S = sin4A + sin4B + sin4C [A+B+C = 180] 

= 2 sin(2A + 2B) · cos(2A – 2B) + 2 sinc.cos2C [2A + 2B + 2C = 360] 

= (-sin2C) · cos(2A – 2B) + (2 sin2C · cos2C) [2A + 2B = 360 – 20] 

= -2 sin2C [cos(2A – 2B) – cos2C] [sin(2A + 2B) = – sin2C] 

= – 2 sin2C [cos(2A – 2B) – cos(2A + 2B)] [cos(2A + 2B) = cos 2C] 

= -2 sin2C[-2 sin2A · sin(-2B)] 

= – 4 sin2A · sin2B · sin2C 

= R.H.S.

sin θ = \(\frac{3}4\)

⇒ cosθ = \(\frac{\sqrt{7}}4\)

\(\sqrt{\frac{cosec^2θ -cot^2θ}{sec^2θ-1}}\)

= \(\sqrt{\frac{1+cot^2θ -cot^2θ}{1+tan^2θ-1}}\)

= \(\sqrt{\frac{1}{tan^2\theta}}\)

= cotθ = \(\frac{cosθ}{sinθ}\)

= \(\frac{\sqrt{7}}4\) = \(\frac{4}3\) = \(\frac{\sqrt{7}}3\)

L.H.S: tan⁴θ + tan² θ

Taking tan² θ common, we get

tan² θ (tan² θ + 1)

= tan² θ sec² θ [∵ sec² θ – tan² θ = 1 ⇒ tan² θ + 1 = sec² θ]

= (sec² θ – 1) sec² θ [∵tan² θ = sec² θ – 1]

= sec⁴ θ – sec² θ

: R.H.S

Given: 2 sin² θ – cos² θ = 2

⇒ 2 sin² θ – (1 – sin² θ) = 2 [∵ sin² θ + cos² θ = 1 ⇒ cos² θ = 1 – sin² θ]

⇒ 2 sin² θ – 1 + sin² θ = 2

⇒ 3 sin² θ = 3

⇒ sin² θ = 1

⇒ sin θ = 1

⇒ θ = sin^-1 (1)

⇒ θ = 90°

Correct option is: D) 1 : √3 : 2

By sine formula for triangle,

\(\frac a {sin\, A} = \frac b{sin\,B} = \frac c{sin \, C} = \lambda\) (Let)

= a = \(\lambda\) sin A, where a is length of side opposite to <A.

b = \(\lambda\) sin B, where b is length of side opposite to < B.

c =  \(\lambda\) sin C, where c is length of side opposite to < C.

Let A = \(30^\circ\), B = \(60^\circ\) & C = \(90^\circ\)

Then a = \(\lambda\) sin \(30^\circ\) = \(\frac \lambda2\)

b = \(\lambda\) sin \(60^\circ\) = \(\frac {\sqrt3 \lambda}2\) 

c = \(\lambda\) sin \(90^\circ\) = \(\lambda\)

\(\therefore\) a : b : c = \(\frac \lambda2\) : \(\frac {\sqrt3 \lambda}2\) : \(\lambda\)

= \(\lambda\) : \(\sqrt3 \lambda\) : 2\(\lambda\) (\(\because\) On multiplying by 2)

= 1 : \(\sqrt3\) : 2 (\(\because\) On multiplying by \(\lambda\))

Hence, the ratio of lengths of opposite side of \(30^\circ\), \(60^\circ\) , \(90^\circ\) is 1 : \(\sqrt3\): 2

Correct option is: D) 1 : √3 : 2

Correct option is: C) 1

\(Sin^2 \theta . Cot^2 \theta + Cos^2 \theta . Tan^2 \theta\)

= \(sin^2\theta.\frac {cos^2\theta}{sin^2\theta} + cos^2\theta. \frac {sin^2\theta}{cos^2\theta}\) (\(\because\) tan \(\theta\) = \(\frac {sin\, \theta}{cos\, \theta}\) & cot \(\theta\) = \(\frac {cos\, \theta}{sin \, \theta}\))

= \(cos^2\theta + sin^2\theta \)

= 1 (\(\because\) \(cos^2\theta + sin^2\theta \) = 1)

Correct option is: C) 1

Correct option is: B) 1

Given that 

x sin (\(90^\circ\) - \(\theta\)) cot(\(90^\circ\) - \(\theta\)) = cos (\(90^\circ\) - \(\theta\))

= x cos \(\theta\) tan \(\theta\) = sin \(\theta\)

(\(\because\) cos (\(90^\circ\) - \(\theta\)) = sin \(\theta\), sin (\(90^\circ\) - \(\theta\)) = cos \(\theta\), cot (\(90^\circ\) - \(\theta\)) = tan \(\theta\))

= x cos \(\theta\) \(\frac {sin \, \theta}{cos \, \theta} = sin \, \theta\) (\(\because\) tan \(\theta\) = \(\frac {sin \, \theta}{cos \, \theta} \))

= x sin \(\theta\) = sin \(\theta\)

= x = \(\frac {sin \, \theta}{sin \, \theta} \) = 1

Correct option is: B) 1

Correct option is: B) 1

Sec \(\theta\) (1 – Sin \(\theta\)) (Sec \(\theta\) + Tan \(\theta\)) 

= sec \(\theta\) (1- sin \(\theta\)) (\(\frac 1{cos\, \theta} + \frac {sin \, \theta}{cos \, \theta}\))

= sec \(\theta\) ( 1- sin \(\theta\)) . \(\frac {1+sin\, \theta}{cos \, \theta}\) 

= \(\frac {1-sin^2\theta}{cos^2\theta} \) (\(\because\) sec \(\theta\) = \(\frac 1{cos\, \theta}\) )

= \(\frac {cos^2\theta}{cos^2\theta} = 1 \)  (\(\because\) \(1- sin^2\theta = cos^2\theta\))

Correct option is: B) 1

Correct option is: B) 1

We have sin\(\theta\) + \(sin^2\theta\) = 1...(1)

= \(1-sin^2\theta = sin\,\theta\)

= \(cos^2\theta = sin\,\theta\)  ...(2) (\(\because\) \(1- sin^2\theta = cos^2\theta\))

Now, \(cos^2\theta + cos^4\theta = cos^2\theta + (cos^2\theta)^2\)

= sin \(\theta\) + \((sin \, \theta)​^2​​​​​​\)(\(\because\) \(cos^2\theta = sin \theta \) from (2))

= sin \(\theta\) + \(sin^2\theta\)

= 1    (From (1))

Correct option is: B) 1

Correct option is: C) 1

\(sec^2 \,27^\circ – cot^2 \,63^\circ\)

= \(1 + tan^2 \, 27^\circ - cot^2\, 63^\circ \) (\(\because\) \(sec^2 \theta = 1 + tan^2\theta\))

= \(1 + tan^2 (90^\circ - 63^\circ) - cot^2\, 63^\circ\) ( \(\because\) \(27^\circ\) = \(90^\circ\) - \(63^\circ\))

= \(1 + cot^2 \, 63^\circ - cot^2\, 63^\circ\) (\(\because\) tan (\(90^\circ\) -\(\theta\)) = cot \(\theta\))

= 1

Correct option is: C) 1

Correct option is: B) -1

\(tan^2 \theta – sec^2 \theta = -(sec^2\theta - tan^2\theta) = -1\)

(\(\because\) \(sec^2\theta - tan^2\theta = 1\))

Correct option is: B) -1

In a ∆ABC right angled at B, ∠A = ∠C, therefore

∠A = ∠C = 45°

(i)

sin A cos C + cos A sin C 

= sin (45° + 45°)

= sin 90°

= 1

(ii)

sin A sin B + cos A cos B 

= cos (A + B)

= cos (90° + 45°)

= sin (45°)

= \(\frac{1}{\sqrt{2}}\)

(sin 65° + cos 25°)(sin 65° − cos 25°) = 0

LHS = (sin 65° + cos 25°)(sin 65° − cos 25°)

= sin² 65° – cos² 25°

= sin² 65° – cos² (90 – 65)°

= sin² 65° – sin² 65°

= 0

= RHS

sin 25° cos 65° + cos 25° sin 65° 

= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°) 

= sin 25° sin 25° + cos 25° + cos 25° 

= sin² 25° + cos² 25° 

= 1. [∵ cos² θ + sin² θ = 1]

Given that sec θ + tan θ = p , 

We know that sec² θ – tan² θ = 1 

sec² θ – tan² θ = (sec θ + tan θ) (sec θ – tan θ) 

= p (sec θ – tan θ) 

= 1 (from given) 

⇒ sec θ – tan θ = \(\frac{1}{p}\)

Given that 

(1 – cos θ) (1 + cos θ) (1 + cot² θ) 

= (1 – cos² θ) (1 + cot² θ) 

[∵ (a – b) (a + b) = a² – b²] 

= sin² θ. cosec² θ [∵ 1 – cos² θ = sin² θ; 1 + cot² θ = cosec² θ] 

= sin² θ . \(\frac{1}{sin^2\, θ}\) [∵ cosec θ = \(\frac{1}{sin\, θ}\) ] 

= 1