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Answer is (A) 3

tan² 60° = (√3)² = 3

Answer is (C) 1/2

cos 45° = (1/√2)² = 1/2

Answer is (B) π/3

cosec θ = 2/√3

cosec 60° = cosec π/3

θ = π/3

Answer is (C) π/4

sec θ = √2

sec θ = sec 45° = sec π/4

θ = π/4

Answer is (D) 1/√2

tan θ = 1

⇒ tan θ = tan 45°

∴ θ = 45°

cos 45° = 1/√2

tan 3x = sin 45° cos 45° + sin 30°

⇒ tan 3x = 1/√2 × 1/√2 + 1/2

⇒ tan 3x = 1/2 + 1/2

⇒ tan 3x = 1

⇒ tan 3x = tan 45°

⇒ 3x = 45°

⇒ x = 15°

Answer is (D) 90°

Given,
sin α = 1/2

⇒ sin α = sin 30°

α =30°

and cos β = 1/2

⇒ cos β = cos 60°

β = 60°

Now, α + β = 30° + 60° = 90°

Answer is (B) 45°

sin θ = cos θ

⇒ sinθ/cosθ = 1

⇒ tan θ = 1

⇒ tan θ = tan 45°

∴ θ = 45°

L.H.S. = \(\frac1{1+sin\,θ}+ \frac1{1-sin\,\theta}\) = \(\frac{1-sin\,\theta+1+sin\,\theta}{(1+sin\,\theta)(1-sin\,\theta)}\)

= \(\frac2{1-sin^2\,θ}= \frac2{cos^2\,\theta}\) = 2sec² θ

R.H.S.= 2 cosec² (90° – θ) = 2 sec² θ. 

∴ L.H.S. = R.H.S

(b) \(\frac{7\sqrt3-6}{6}.\)

Given exp. = \(0+\frac{\sqrt3}{2}-1+\frac{2}{\sqrt3}+0\)

= \(\frac{\sqrt3}{2}-1+\frac2{\sqrt3}=\frac{2\sqrt3+4}{2\sqrt3}=\frac{7-2\sqrt3}{2\sqrt3}\times\frac{\sqrt3}{\sqrt3}\)

= \(\frac{7\sqrt3-6}{6}.\)

L.H.S. = cos² 0° – 2 cot² 30° + 3 cosec² 90°

= (1)² – 2(√3)² + 3(1)²

= 1 – 2 × 3 + 3

= 1 – 6 + 3 = -2

R.H.S. = 2(sec² 45° – tan² 60°)

= 2[(√2)² – (√3)²]

= 2(2 – 3) = 2 × (-1) = -2

Thus, L.H.S. = R.H.S.

4 cot² 45° – sec² 60° – sin² 30° = – 1/4

L.H.S. = 4 cot² 45° – sec² 60° – sin² 30°

= 4(1)² – (2)² – \({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\)

= 4 – 4 – 1/4 = –1/4

= R.H.S.

Hence, L.H.S. = R.H.S.

RHS = 2 cos² 30° – 1

= 2(√3/2)² - 1

= 2 x 3/4 - 1

= 3/2 - 1 = 1/2

L.H.S. = cos 60° = 1/2

∴ L.H.S. = R.H.S.

(i) A = B = 60° (Given)

LH.S. cos(A – B) = cos(60° – 60°)

= cos 0° = 1

R.HS. cosA cosB + sinA sinB

= cos 60° cos 60° + sin 60° sin 60°

cos² 60° + sin² 60° = 1 [ sin² θ + cos² θ = 1]

Thus, L.H.S. R.H.S.

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

L.H.S. sin(A – B) = sin(60° – 60°)

= sin 0° = 0

R.H.S. sin A cos B – cos A sin B

= sin 60° cos 60° – cos 60° sin 60°

= 0

Thus, L.H.S. = R.H.S.

cosec² 30° – 3 sec 60°

= (2)² – 3(2)

= 4 – 3 × 2 = 4 – 6

= -2

sin² 60° cot² 60°

= (√ 3/2)² x (1/√ 3)²

= 3/4 x 1/3

= 1/4

4 cos³ 30° – 3 cos 30°

= 4 x (√3/2)³ - 3 x √3/2

= 4 x 3√3/8 - 3√3/2

= 3√3/2 - 3√3/2 = 0

Answer is (B) 0

sin (45° + θ) – cos (45° – θ)

= sin (45° + θ) – cos [90° – (45° + θ)] [∵ (45° – θ) = {90° – (45° + θ)}]

= sin (45° + θ) – sin (45° + θ) [∵ cos (90° – θ) = sin θ]

= 0

(d) 1

sin² 25° + sin² 65° = sin² (90° – 65°) + sin² 65° 

= cos² 65° + sin² 65° = 1.

(b) θ – ϕ = 0

sin (30° – θ) = cos (60° + ϕ) 

⇒ cos (90° – (30° – θ) = cos (60° + ϕ)           (∵ sin θ = cos (90° – θ)) 

⇒ cos (60° + θ) = cos (60° + ϕ) 

⇒ 60° + θ = 60° + ϕ ⇒ θ – ϕ = 0.

(b) 0

tan θ = 1 ⇒ tan θ = tan 45° ⇒ θ = 45° 

sin ϕ = \(\frac1{\sqrt2}\)⇒ sin ϕ = sin 45° ⇒ ϕ = 45° 

∴ cos (θ + ϕ) = cos (45° + 45°) 

= cos 90° = 0

(d) 1

Given exp. = cos² θ [1+ tan² θ] 

= cos² θ . sec² θ = 1

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

(a) 30°

cos² θ – sin² θ = \(\frac12\)

⇒ (1–sin² θ) – sin² θ = \(\frac12\)

⇒ 1– 2 sin2 θ = \(\frac12\)

⇒ 2 sin² θ = \(\frac12\) ⇒ sin² θ = \(\frac14\)

⇒ sin θ = \(\frac12\) = sin 30° ⇒ θ = 30°

(c) 2

Given exp.

= \(\frac{sin(90°-71°)}{cos\,71°}+\frac{cox\,73°}{sin(90°-73°)}\)

= \(\frac{cos\,71°}{cos\,71°}+\frac{cos\,73°}{cos\,73°}\)            (∵sin (90° – θ) = cos θ)

= 1 + 1 = 2.

(c) 0

tan 26° – cot 64° = tan (90° – 64°) – cot 64° 

= cot 64° – cot 64° = 0 (∵ tan (90° – θ)= cot θ)

(b) 15°

sin 2x = \(\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}-\frac12\times\frac12\)

= \(\frac34-\frac14 = \frac12\)

∴  sin 2x = \(\frac12\) = sin 30° ⇒ 2x = 30° ⇒ x = 15°. 

(c) \(\frac{47}{24}\)

Given exp.

= \(\bigg(\frac12\bigg)^2\times\bigg(\frac1{\sqrt2}\bigg)^2+4\times\bigg(\frac1{\sqrt3}\bigg)^2+\frac12\times(1)^2-2\times0\)

= \(\frac14\times\frac12+4\times\frac13+\frac12=\frac18+\frac43+\frac12=\frac{3+32+12}{24}=\frac{47}{24}\)

  \(\frac{cos\,70°}{sin\,20°}\)+ cos 49°cosec 41° = \(\frac{cos(90°-20°)}{sin\,20°}+\)cos(90° - 41°)cosec 41°

= \(\frac{sin\,20°}{sin\,20°}+\) sin 41° x \(\frac1{sin\,41°}\) = 1 + 1 = 2.

Tan 3\(x\) = sin 45° cos 45° + sin 30°

= \(\frac{1}{\sqrt2}\times\frac{1}{\sqrt2}+\frac12\) = \(\frac12+\frac12+1\)

⇒ tan 3\(x\) = tan 45° ⇒ 3\(x\) = 45° ⇒ \(x\) = 15°

(b) sin x

\(x\) + y = 90° ⇒ y = 90° – \(x\)

∴ \(\sqrt{\text{cos}\,x\,cosec\,y\,-\,cos\,x\,sin\,y}\)

= \(\sqrt{cox\,x\,cosec\,(90°-x)-cox\,x\,sin(90°-x)}\)

= \(\sqrt{cos\,x\,sec\,x\,-cos\,x\,cos\,x}\)

= \(\sqrt{1-cos^2\,x}=\sqrt{sin^2\,x}=sin\,x\)

Given exp = \(\frac43\times\bigg(\frac{1}{\sqrt3}\bigg)^2+\bigg(\frac{\sqrt3}{2}\bigg)^2-3\times\big(\frac12\big)^2+\frac34\times(\sqrt3)^2-2\times(1)^2\)

= \(\frac43\times\frac13+\frac34-\frac34+\frac94-2\)

= \(\frac49+\frac94-2=\frac{16+81-72}{36}= \frac{25}{36}.\)

(d) 2

\(x\) cos 60° + y cos θ° = 3

⇒ \(x\) x \(\frac12\) + y x 1 = 3

⇒ \(x\) + 2y = 6                     ...(i)

and 4\(x\) sin 30° – y cot 45° = 2

⇒ 4\(x\) x \(\frac12\) - y x 1 = 2

⇒ 2\(x\) – y = 2                  ...(ii) 

Now solve for x and y

(c)  \(\frac1{\sqrt3}\)

2 sin² \(x\) + cos² 45° = tan 45° 

⇒ 2 sin² \(x\) + \(\bigg(\frac1{\sqrt2}\bigg)^2\) = 1 

⇒ 2 sin² \(x\)  =  \(1-\frac12=\frac12\)

 ⇒ sin² \(x\) = \(\frac14\) ⇒ sin \(x\) = \(\frac12\) = sin 30° ⇒ \(x\) = 30°

∴ tan \(x\) = tan 30° = \(\frac1{\sqrt3}\).

(d) c

a sin 0° + b cos 90° + c tan 45° 

= a × 0 + b × 0 + c × 1 = c.

(d) 0 

\(\frac{1-tan^2\,45°}{1+tan^2\,45°}\) = \(\frac{1-1}{1+1}= \frac02=0\)            (∵ tan 45° = 1)

(a) \(\frac23\)

 \(x\) tan 30° = \(\frac{\text{sin 30° + cos 60°}}{\text{tan 60° + sin 60°}}\)

= \(\frac{\frac12+\frac12}{\sqrt3+\frac{\sqrt3}{2}}=\frac{1}{\frac{2\sqrt3+\sqrt3}{2}}=\frac{2}{3\sqrt3}\)

\(x\times\frac1{\sqrt3}=\frac{2}{3\sqrt3}\) ⇒ \(x\) = \(\frac{2\sqrt3}{3\sqrt3}=\frac23.\)

(b) 23°

sin 3θ = cos (θ – 2°) 

⇒ cos (90° – 3θ ) = cos (θ – 2°) 

⇒ 90° – 3θ = θ – 2° 

⇒ 4θ = 92° ⇒ θ = 23°.