Explore topic-wise InterviewSolutions in .

Correct answer is

(C) sec²θ

Correct answer is

(B) √2

Correct answer is

(A) 1

Correct option is: A) \(\frac{9}{2}\)

\(sin^2 \, 60^\circ + cos^2 \, 30^\circ + tan^2\, 60^\circ\)

= \((\frac {\sqrt3}2)^2 + (\frac {\sqrt3}2)^2 + ({\sqrt3})^2\)  (\(\because\) sin \(60^\circ\) = \(\frac {\sqrt3}2\), cos \(30^\circ\) =  \(\frac {\sqrt3}2\) and tan \(60^\circ\) = \(\sqrt3\))

= \(\frac 34 + \frac 34 + 3\)

= \(\frac 64 + \frac {12}4 = \frac {18}4 = \frac 92\)

Hence, \(sin^2\, 60^\circ + cos^2\, 30^\circ + tan^2\, 60^\circ = \frac 92\)

Correct option is: A) \(\frac{9}{2}\)

Correct option is: B) \(\frac{7}{17}\)

Given that

8 tan A = 15

= tan A = \(\frac {15}8\)

\(\because\) \(sec^2 A = 1 + tan^2 A\)

= \(1 + (\frac {15}{8})^2\)

= 1 + \(\frac {225}{64}\)

= \(\frac {64 + 225}{64} = \frac {289}{64}\)

= \((\frac {17}{8})^2\)

\(\therefore\) sec A = \(\frac {17}8\)

\(\therefore\) cos A = \(\frac 1 {sec \, A} = \frac 1{\frac {17}8} = \frac 8{17}\)

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

= sin A = cos A. tan A = \(\frac 8 {17} \times \frac {15}8 = \frac {15}{17}\)

Now, sin A - cos A = \( \frac {15}{17} - \frac 8 {17} = \frac {15-8}{17} = \frac 7{17}\)

Correct option is: B) \(\frac{7}{17}\)

(sin x – cos x)² + (sin x + cos x)² 

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

= 2(sin² x + cos² x) = 2(1) = 2

Correct option is: D) √2

\(\sqrt{1 + sin^2\, \theta + cos^2\, \theta} = \sqrt {1+1} = \sqrt2\) (\(\because\) \(sin^2\theta + cos^2\theta = 1\))

Correct option is: D) √2

Correct option is: C) 22.5

We have tan 2 A = 1 = tan \(45^\circ\)

= 2A = \(45^\circ\)

= A = \(\frac {45^\circ}{2} = (22.5)^\circ\)

Correct option is: C) 22.5

Correct option is: D) \(1 + \frac{z^2}{c^2}\)

We have

x = a sec \(\theta\) cos \(\phi\)  = \(\frac xa\) sec \(\theta\) cos \(\phi\).

y = b sec \(\theta\) sin \(\phi\) = \(\frac yb\) = sec \(\theta\) sin \(\phi\).

z = c tan \(\theta\) 

Now, \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = (\frac xa)^2 + (\frac yb)^2\)

= \((sec\, \theta \, cosc \, \phi)^2 + (sec\, \theta \, sin\, \phi)^2\)

= \(sec^2\theta \, cos^2\phi + sec^2\theta \, sin^2\phi\)

= \(sec^2\theta (cos^2\, \phi + sin^2\phi)\)

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

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

= 1+ \((\frac zc)^2\) (\(\because\) z = c \(\tan \theta = tan \theta = \frac zc\))

= 1 + \(\frac {z^2}{c^2}\)

Correct option is: D) 1 + \(\frac{z^2}{c^2}\)

(i) sin12x + sin4x = 2.sin \(\frac{12x + 4x}{2}.cos\frac{12x - 4x}{2}\). cos 

(ii) sin7A – sin3A = 2cos5A.sin2A 

(iii) cos6θ + cos2θ = 2 cos4θ.cos2θ 

(iv) sin 80° – sin 40° = 2 cos60°. sin 20° = 2. sin 20° = sin 20°

 L.H.S = \(\frac{sin\,x-sin\,3x+sin\,5x-sin\,7x}{cos\,x-cos\,3x-cos\,5x+cos\,7x}\)

= \(\frac{(5sin\,x + sin\,x)-(sin\,7x+sin\,3x)}{(cos\,x-cos5x)-(cos\,3x-cos\,7x)}\)

= \(\frac{2sin(\frac{5x+x}{2}).cos(\frac{5x-x}{2})-2sin(\frac{7x+3x}{2}).cos(\frac{7x-3x}{2})}{2sin(\frac{x+5x)}{2}).sin(\frac{5x-x}{2})-2sin(\frac{3x+7x}{2}).sin(\frac{7x-3x}{2})}\)

=\(\frac{2\,sin\,3x\,.cos\,2x-2\,sin\,5x.cos\,2x}{2\,sin\,3x.sin\,2x-2sin\,5x.sin\,2x}\)

= \(\frac{2\,cos\,2x(sin\,3x-sin\,5x)}{2\,sin\,2x(sin\,3x-sin\,\,5x)}\)

= \(\frac{cos\,2x}{sin\,2x}=cot\, 2x = R.H.S.\)

(a) 1

\(\bigg(\frac{x}{a}\bigg)^\frac23 +\bigg(\frac{y}{b}\bigg)^\frac23\) = \(\bigg(\sqrt[3]{\frac{a\,cos^3\,\theta}{a}}\bigg)^2 + \bigg(\sqrt[3]\frac{b\,sin^3\,\theta}{b}\bigg)^2\)

= (sin θ)² + (cos θ)² = sin² θ + cos² θ = 1

(b) a² – b²

x² – y² = (a sec θ + b tan θ)² – (b sec θ + a tan θ)² 

= a² sec² θ + 2ab sec θ tan θ + b² tan² θ - (b² sec² θ + 2ab sec θ tan θ + a² tan² θ)

= (a² - b²) sec² θ + (b² - a²) tan² θ

= (a² - b²)(sec² θ - tan² θ)

[(∴ a² - b² = - (b² - a²))(∵ sec² θ - tan² θ = 1)]

= a² – b²

(c) 1

sin² θ cos² θ (sec² θ + cosec² θ)

= sin² θ cos² θ \(\bigg(\frac{1}{cos^2\,\theta}+\frac{1}{sin^2\,\theta}\bigg)\)

 = sin² θ cos² θ \(\bigg(\frac{sin^2\,\theta\,+\,cos^2\,\theta}{cos^2\,\theta\,sin^2\,\theta}\bigg)\)

= sin² θ + cos² θ = 1

(c) 

7 sin A = 24 cos A

⇒ \(\frac{sin\,A}{cos\,A}=\frac{24}{7}\) ⇒ tan A = \(\frac{24}{7}\)

Now make the figure and find all the other t-ratios and solve.

L.H.S.

= cos (3π/2 +x) cos (2π+x) .[cot(3π/2-x)+ cot (2π+x)] 

= (sin x)(cos x) (tan x + cot x)

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

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

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

= 1 = R.H.S

Correct option is: B) 105°

We have 

cos (A-B) =  \(\frac 12\) = cos \(60^\circ\)

= A - B = \(60^\circ\)....(1)

And sin B = \(\frac 1{\sqrt2}\) = sin \(45^\circ\)

= B = \(45^\circ\)

\(\therefore\) A- \(45^\circ\) = \(60^\circ\) (From (1))

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

Correct option is: B) 105°

Given function is : tan (C + A)/2 = cot B/2

Sum of all the angles of a triangle = 180 degree

So, A + B + C = 180°

Or A + C = 180° – B

And, (A + C)/2 = (180° – B)/2 = 90° – B/2

Now, tan (A + C)/2 = tan(90° – B/2) = cot B/2

Hence Proved.

cosec 54° + sin 72° 

= cosec (90 – 36)° + sin (90 – 18)° 

= sec 36° + cos 18°

sec 76° + cosec 52° 

= sec (90 – 14)° + cosec (90 – 38)° 

= cosec 14° + sec 38°

sec A = (1/cos A) = (1/√(1 - sin² A))

and

tan A = (sin A/cos A) = (sin A/√(1-sin² A))

Correct option is: C) 0

cos \(12^\circ\) - sin \(78^\circ\) = cos \(12^\circ\) - sin (\(90^\circ\)- \(12^\circ\))

= cos \(12^\circ\) cos \(12^\circ\) (\(\because\) sin (\(\because\) \(90^\circ\) - \(\theta\)) = cos \(\theta\) =0

Correct option is: C) 0

 Correct option is: D) 0°

We have sin (A + B) = \(\frac 1{\sqrt2} = sin \, 45^\circ\)

= A + B = \(45^\circ\) .....(1) and 

cos (A-B) = \(\frac 1{\sqrt2} = cos\, 45^\circ\)

= A - B = \(45^\circ\) .....(2)

Subtract equation (1) from (2), we get

(A + B) - (A-B) = \(45^\circ\) - \(45^\circ\)

= 2B = \(0^\circ\)

= B = \(0^\circ\)

Hence, < B = \(0^\circ\)

Correct option is: D) 0°

Correct answer is (b) 3

Sec² 60° − 1 

= (2)² − 1 

= 4 − 1 

= 3

Given that cos 12° – sin 78° 

= cos 12° – sin(90° – 12°) [∵ sin (90 – θ) = cos θ] 

= cos 12° – cos 12° = 0

(sin A + cos A) sec A

= (sin A + cos A) 1/cos A 

= sin A/cos A + cos A/cos A 

= tan A + 1 

= 5/12 + 1/1 

= 5+12/12 

= 17/12

We know that cosine function is periodic with period 2π. 

cos 1140° = cos (3 × 360° + 60°) 

= cos 60° = 1/2

sin² 30° cos² 45° + 4 tan² 30° + \(\frac{1}2\) sin² 90° - 2 cos² 90° + \(\frac{1}{24}\) cos² 0°

= \((\frac{1}2)^2\) x \((\frac{1}{\sqrt2})^2\) + 4 x \((\frac{1}{\sqrt3})^2\) + \(\frac{1}2\) x (1)² - 2 x (0)² + \(\frac{1}{24}\) x (1)² 

= \(\frac{1}4\)x \(\frac{1}2\) + \(\frac{4}3\) + \(\frac{1}2\) - 0 = \(\frac{1}{24}\)

= \(\frac{1}8\) + \(\frac{4}3\) + \(\frac{1}2\) + \(\frac{1}{24}\) = 2

Solution: tan (A + B) = √3

⇒ tan (A + B) = tan 60°

⇒ (A + B) = 60° ... (i)

 tan (A – B) = 1/√3

⇒ tan (A - B) = tan 30°
⇒ (A - B) = 30° ... (ii)

Adding (i) and (ii), we get

A + B + A - B = 60° + 30°

2A = 90°

A= 45°

Putting the value of A in equation (i)

45° + B = 60°

⇒ B = 60° - 45°

⇒ B = 15°

Thus, A = 45° and B = 15°  

tan (A + B) = √3 ⇒ A + B = 60° ………….. (i) 

tan (A – B) = \(\frac{1}{\sqrt3}\) ⇒ A – B = 30° ………… (ii) 

From Eqn. (i) + Eqn.(ii), we have 

2A = 90 ⇒ = 45° 

From eqn. (i), A + B = 60° 

45° + B = 60° 

∴ B = 15

Correct option is: D) 15°

We have tan (B + \(15^\circ\) ) = \(\frac 1{\sqrt3} \) = tan \(30^\circ\)

= B + \(15^\circ\) = \(30^\circ\)

= B = \(30^\circ\) - \(15^\circ\) = \(15^\circ\)

Correct option is: D) 15°

Correct option is: D) 45°

We have sin (A-B) = \(\frac 12 = sin \, 30^\circ\)

= A - B = \(30^\circ\) ...(1)

and cos (A+B) \(\frac 12\) = cos  \(60^\circ\) 

= A + B = \(60^\circ\).....(2)

By adding equations (1) & (2), we get 

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

= 2A = \(90^\circ\)

= A = \(\frac {90^\circ}2 = 45^\circ\)

Hence, \(\angle\)A = \(45^\circ\)

Correct option is: D) 45°

Correct option is: C) 0

cos \(45^\circ\). cos \(30^\circ\). cos \(90^\circ\). cos \(60^\circ\) = \(\frac {1}{\sqrt2}. \frac {\sqrt3}2. 0. \frac 12 = 0\)

Correct option is: C) 0

 \(\frac{4}{cot^230°}\) + \(\frac{1}{cot^260°}\) - cos² 45° = \(\frac{4}{(\sqrt3)^2}\) + \(\frac{1}{(\frac{\sqrt3}{2})^2}\) - \((\frac{1}{\sqrt2})^2\)

= \(\frac{4}3\) + \(\frac{4}3\) - \(\frac{1}2\) = \(\frac{13}6\)

 \(\frac{tan45°}{cosec30°}\) +\(\frac{ sec60°}{cot45°}\) - \(\frac{5sin90°}{2cos0°}\)

= \(\frac{1}{2} + \frac{2}{1} - \frac{5\times 1}{2 \times 1}\)

= \(\frac{1}{2} + 2 - \frac{5}{2}\)

= 0

(cosec² 45° sec² 30°) (sin² 30° + 4 cot² 45° - sec² 60°)

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

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

cos² 30° + cos² 45° + cos² 60° + cos² 90°

= \((\frac{\sqrt{3}}{2})^2 \) + \((\frac{1}{\sqrt{2}})^2 \) + \((\frac{1}{2})^2 \) + (0)²

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

Answer : (d) \(\frac{(2r+1)π}{2(m-n)}\)  

tan mθ + cot nθ = 0 ⇒ tan mθ = – cot nθ

⇒ tan mθ = tan (π/2 + nθ) (∵ tan (π/2 + x) = – cot x) 

⇒ mθ = rπ + π/2 + nθ, r∈I 

⇒ (m – n)θ = (2r + 1) π/2, r∈I

⇒ θ = \(\frac{(2r+1)π}{2(m-n)}\) , r \(\in\) I

Taking LHS

= 2(sin⁶ θ – cos⁶ θ) – 3(sin⁴ θ+ cos⁴ θ ) + (sin² θ + cos² θ)

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

Now, we use these identities, (a³ – b³)= (a + b)³ – 3ab(a + b) and (a² + b²) = (a +b)² – 2ab]

= 2[(sin² θ + cos² θ)³ – 3sin²θ cos²θ (sin² θ+ cos² θ)] –3[(sin² θ + cos² θ) – 2 sin² θ cos² θ]+ 1

=2[(1) – 3sin²θ cos²θ (1)] – 3[(1) – 2 sin² θ cos² θ] + 1 [∵ cos² θ + sin² θ = 1]

=2(1 – 3 sin² θ cos² θ )– 3 + 6sin² θ cos² θ+ 1

= 2– 6sin²θ cos²θ – 2 + 6sin² θ cos² θ

= 0

= RHS

Hence Proved

L.H.S. = 16 sin θ cos θ cos 2θ cos 4θ cos 8θ 

= 8(2sinθ cosθ) cos2θ cos 4θ cos 8θ 

= 8sin 2θ cos 2θ cos 4θ cos 8θ 

= 4(2sin 2θ cos 2θ) cos 4θ cos 8θ

= 4sin 4θ cos 4θ cos 8θ = 2(2sin 4θ cos 4θ) cos 8θ = 2sin 8θ cos 8θ = sin 16θ = R.H.S.

Taking LHS

= sin⁸ θ – cos⁸ θ

= (sin⁴ θ)² – (cos⁴ θ)²

= (sin⁴ θ – cos⁴ θ)(sin⁴ θ+ cos⁴ θ )

[∵ (a² – b²) = (a + b) (a – b)]

= {(sin² θ)² – (cos² θ)²}{(sin² θ)² + (cos² θ)²}

= (sin² θ + cos² θ) (sin² θ – cos² θ) [(sin² θ + cos² θ) – 2 sin² θ cos² θ]

[ ∵ (a² + b²) = (a +b)² – 2ab]

= (1)[ sin² θ –cos² θ][(1) – 2 sin² θ cos² θ]

= (sin² θ – cos² θ)(1 – 2 sin² θ cos² θ)

= RHS

Hence Proved