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Given that tan A = cot B 

⇒ cot (90° – A) = cot B [∵ tan θ = cot (90 – θ)] 

⇒ 90° – A = B 

⇒ A + B = 90°

tan A = cot B 

tan A = tan (90 – B) 

A = 90 – B 

∴ A + B = 90°.

Given that tan 2A = cot (A – 18°)

⇒ cot (90° – 2A) = cot (A – 18°) [∵ tan θ = cot (90 – θ)] 

⇒ 90° – 2A = A – 18° 

⇒ 108° = 3A 

⇒ A =  108° / 3  = 36° 

Hence the value of A is 36°.

tan 2A = cot (A – 18°) 

cot (90 – 2A) = cot (A – 18) 

90 – 2A = A – 18 

3A = 108° 

∴ A = 36°

Correct option is : (A) -24/25

25 cos² θ + 5 cos θ – 12 = 0 

∴ (5cos θ + 4) (5 cos θ – 3) = 0 

∴ cos θ = -4/5 or cos θ = 3/5

Since π/2 < α < π,

cos α < 0

∴ cos α = -4/5

sin² α = 1 – cos² α = 1 – 16/25 = 9/25

∴ sin α = ± 3/5

Since π/2 < α < π sin α > 0

∴ sin α = 3/5 

sin 2 α = 2 sin α cos α

= 2 (3/5) (-4/5) = -24/25

We know that

cos θ = sin (90° - θ)

= sin (90° - 38°) sin (90° -52°) – sin 38° sin 52°

= sin 52° sin 38°– sin 38° sin 52°

= 0

We know that

tan (60°) = √3

cosec (45°) = √2

sec (60°) = 2

tan(45°) = 1

Now putting the values;

= (√3) × (√2)² + (2)² × (1)

= 2√3 + 4

=2 (√3 + 2)

Correct answer is (a) 30°

2 sin 2θ = √3 

⇒ sin 2θ = √3/2 = sin 60°

⇒ sin 2θ = sin 60°

⇒ 2θ = 60°

⇒ θ = 30°

cot 85° + cos 75° = cot (90° – 5°) + cos (90° – 15°)

= tan 5° + sin 15°

sin 67° + cos 75° 

= sin (90 – 23) + cos (90 – 15) 

= cos 23° + sin 15°.

Given: cos 75° + cos 75°

To find : Expression in terms of angles between 0° and 30°. 

Solution : Use the values:

cos(90° - θ) = sinθ

cot (90° - θ) = tanθ

Solve,

cot 75° + cot 75° 

=  cos (90° - 15°) + cot(90° - 15°)

= sin 15° + tan 15°

Here, sin (A – B) = 1/2 

⇒ sin (A – B) = 30° [∵ sin 30° = 1/2 ] 

⇒ (A – B) = 30° …….(i) 

Also, cos (A + B) = 1/2 

⇒ cos (A + B) = cos 60° [∵ cos 60° = 1/2 ] 

⇒ A + B = 60° ….(ii) 

Solving (i) and (ii), we get: 

A = 45° and B = 15°

Here, tan (A – B) = 1/√3 

⇒ tan (A – B) = tan 30° [∵ tan 30° = 1 √3 ] 

⇒ (A – B) = 30° …….(i) 

Also, tan (A + B) = √3

⇒ tan (A + B) = tan 60° [∵ tan 60° =√3] 

⇒ A + B = 60° …….(ii) 

Solving (i) and (ii), we get: 

A = 45° and B = 15°

Taking LHS = sin² θ – cos² φ

=( 1 – cos² θ) – (1 – sin² φ) [∵ cos² θ + sin² θ = 1] & [∵ cos² φ + sin² φ = 1]

= 1 – cos² θ – 1 + sin² φ

= sin² φ – cos² θ

= RHS

Hence Proved

(b) cos 2(θ + ϕ)

Given cos 2θ . cos 2ϕ + sin² (θ – ϕ) – sin² (θ + ϕ) 

= cos 2θ cos 2ϕ + sin (θ – ϕ + θ + ϕ) sin (θ – ϕ – θ – ϕ) 

= cos 2θ cos 2ϕ + sin 2θ sin(-2ϕ)

= cos 2θ cos 2ϕ – sin 2θ sin(2ϕ) 

= cos (2θ + 2ϕ) = cos 2(θ + ϕ)

(i) LHS = (cos 20° + cos 100°) + cos 140°

= 2 cos (\(\frac{20°+100°}{2}\)) + cos (\(\frac{20°-100°}{2}\))140°

[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]

= 2 cos 60° cos(-40°) + cos 140°

= 2 x \(\frac{1}{2}\) x cos(-40°) + cos(180° – 140°)

[∵ cos(-θ) = cos θ, cos 60° = \(\frac{1}{2}\)

= cos 40° – cos 40°

= 0

Hence Proved.

(ii) LHS = (sin 50° – sin 70°) + sin 10°

= 2 cos (\(\frac{50+70}{2}\)) sin  (\(\frac{50+70}{2}\)) + sin 10°

[∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))]

= 2 cos 60° sin(-10°) + sin 10°

= 2 x \(\frac{1}{2}\)(-sin 10°) + sin 10° [∵ sin(-θ) = -sin θ]

= -sin 10° + sin 10°

= 0

= RHS

(a) 0

LHS = (cos 10 + cos 179°) + (cos 2° ÷ cos 178°) + ….. + cos(89° + cos 91°) + cos 90° 

cos 179° = cos (180° – 1) = – cos 1° 

cos 178° = cos(180° – 2) = – cos 2° 

So (cos 1°- cos 1°) + (cos 2° – cos 2°) + (cos 89° – cos 89°) + cos 90° 

= 0 + 0 …. + 0 + 0 = 0.

(i) LHS = (sin 2A + sin 2B) + sin 2C 

= 2 sin (A + B) cos (A – B) + 2 sin C cos C = 2 sin

(90° – C) cos (A – B) + 2 sin C cos C 

= 2 cos C [cos (A – B) + sin C] + cos (A + B) (∴ A + B = π/2 – C) 

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

= 2 cos C [2 cos A cos B] 

= 4 cos A cos B cos C = RHS

(ii) LHS = (cos 2A + cos 2B) + cos 2C 

= 2 cos (A + B) cos (A – B) + 1 – 2 sin² C 

= 1 + 2 sin C (cos (A – B) – 2 sin² C) 

{∴ cos (A + B) = cos (90° – C) = sin C} 

= 1 + 2 sin C [cos (A- B) – sin C] 

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

= 1 + 2 sin C [2 sin A sin B] 

= 1 + 4 sin A sin B sin C = RHS

(a) is an equilateral triangle with side 4m

A triangle will have a max area (with a given perimeter) when it is an equilateral triangle.

(d) 3 sin (α - β)

Given, tan α = 2 tan β ⇒ \(\frac{\frac{sin\,α}{cos\,α}}{\frac{sin\,β}{cos\,β}}=2\)

= \(\frac{sin\,α\,cos\,β}{cos\,α\,sin\,β}=\frac21\)

Using componendo and dividendo, we get

\(\frac{sin\,α\,cos\,β+cos\,α\,sin\,β}{sin\,α\,cos\,β\,-cos\,α\,sin\,β}=\frac{2+1}{2-1}\)

= \(\frac{sin\,(α+β)}{sin\,(α-β)}=\frac31\) ⇒ sin (α + β) = 3 sin (α - β).

Taking LHS = (1 – sinθ)(1+ sinθ)

Using identity , (a + b) (a – b) = (a² – b²) , we get

= (1)² – (sinθ)²

= 1 – sin² θ

= cos² θ [∵ cos² θ + sin² θ = 1]

= RHS

Hence Proved

Taking LHS = (sin θ – cos θ)²

Using the identity,(a – b)² = (a² + b² – 2ab)

= sin² θ + cos² θ – 2sin θ cos θ

= 1 – 2sin θ cos θ [∵ cos² θ + sin² θ = 1]

= RHS

Hence Proved

Correct option is: A) \(\frac{x^2-y^2}{x^2+y^2}\)

Taking RHS =x² + y²

Putting the values of x and y, we get

(a cos θ – b sin θ)² + (a sin θ + b cos θ)²

= a²cos²θ + b² sin²θ – 2ab cos θ sin θ + a²sin²θ + b² cos²θ + 2ab cos θ sin θ

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

= a² + b² [∵ cos² θ + sin² θ = 1]

= RHS

Hence Proved

Taking LHS = (sin θ + cos θ)² + (sin θ – cos θ)²

Using the identity,(a + b)² = (a² + b² + 2ab) and (a – b)² = (a² + b² – 2ab)

= sin² θ + cos² θ + 2sin θ cos θ + sin² θ + cos² θ – 2sin θ cos θ

= 1 +1 [∵ cos² θ + sin² θ = 1]

= 2

= RHS

Hence Proved

sin 59° + cos 56° 

= sin (90 – 31)° + cos (90 – 34)° [sin (90 – θ) = cos θ, cos (90 – θ) = sin θ]

= cos 31° + sin 34°

L.H.S. = tan x + cot x

= \(\frac {sin\,x}{cos\,x} + \frac {cos\,x}{sin\, x}\)

= \(\frac {sin^2x+cos^2x}{sin\,x\,cos\,x}\)

= \(\frac{1}{sin\,x\,cos\,x}=\frac{2}{2sin\,x\,cos\,x}\)

=\(\frac{1}{sin\,2x}\)

= 2 cosec 2x = R.H.S

L.H.S. = (cos x – cos y)² + (sin x – sin y)²

= cos² x + cos² y + 2cos x. cos y + sin² x + sin² y + 2sin x. sin y 

= (cos² x + sin² x) + (cos² y + sin² y) – 2(cos x. cos y + sin x. sin y) 

= 1 + 1 – 2cos(x – y) 

= 2 – 2 cos (x – y) 

= 2[1 – cos(x – y)]

= 2[2sin²[((\(\frac{x-y}{2}\)))]… [∵ 1 – cos θ = 2 sin² θ/2]

= 4 sin²(\(\frac{x-y}{2}\))

=R.H.S

L.H.S. = (cos x + cos y)² + (sin x + sin y)² 

= cos² x + cos² y + 2cos x.cos y + sin² x + sin² y + 2sin x.sin y 

= (cos² x + sin² x) + (cos² y + sin² y) + 2(cos x.cos y + sin x.sin y) 

= 1 + 1 +2cos(x – y) 

= 2 + 2 cos (x – y) 

= 2[1 + cos(x – y)]

= 2[2cos² [((\(\frac{x-y}{2}\)))]...[∵ 1 + cos θ = 2 cos² θ/2]

= 4 cos² (\(\frac{x-y}{2}\))

=R.H.S

L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x 

= sin 3x sin x + sin2 x + cos 3x cos x – cos² x 

= (cos 3x cos x + sin 3x sin x) 

— (cos² x — sin² x) 

= cos (3x – x) – cos 2x 

= cos 2x – cos 2x 

= 0 

= R.H.S.

L.H.S. = cos θ + sin (270° + θ) – sin (270° – θ) + cos (180° + θ) 

= cos θ + (- cos θ)-(- cos θ) – cos θ 

= cos θ – cos θ + cos θ – cos θ 

= 0 

= R.H.S.