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Given

sin⁴ x + cos⁴ x = sin x cos x

⇒ (sin² x + cos² x) ² – 2 sin² x cos² x = sin x cos x 

⇒ 1 - \(\frac{(2\,sinx\,cosx)^2}{2}\) = \(\frac{2\,sinx\,cosx}{2}\)

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

⇒ sin² 2x + sin 2x – 2 = 0 

⇒ (sin 2x + 2) (sin 2x – 1) = 0 

⇒ sin 2x = 1

⇒ sin 2x = sin \(\frac{\pi}{2}\) = sin \(\big(\) 2π + \(\frac{\pi}{2}\) \(\big)\) ( ∵ sin 2x ≠ – 2 is indivisible)

⇒ 2x = \(\frac{\pi}{2}\) or  \(\big(\) 2π + \(\frac{\pi}{2}\) \(\big)\) 

⇒ 2x = \(\frac{\pi}{2}\) or \(\frac{5\pi}{2}\) 

⇒ x = \(\frac{\pi}{4}\) or \(\frac{5\pi}{4}\)

sin x = \(-\frac{\sqrt{3}}{2}\)= – sin 60° 

= sin (180° + 60°) 

= sin (360° – 60°)

⇒ x = 240°, 300°.

sin 3 A = cos (A − 26°) (given)

or cos (90° – 3A) = cos (A – 26°)

On comparing

90° – 3A = A – 26°

A + 3A = 90° + 26°

4A = 116° = 29°

The value of A is 29°.

cos² θ – sin θ - \(\frac{1}{4}\) = 0 

⇒ 1 – sin² θ – sin θ – \(\frac{1}{4}\)  = 0 

⇒ 4 sin² θ + 4 sin θ – 3 = 0 

⇒ (2 sin θ + 3) (2 sin θ – 1) = 0 

⇒ 2 sin θ + 3 = 0 or 2 sin θ – 1 = 0 

⇒ sin θ = - \(\frac{3}{2}\) or sin θ = \(\frac{1}{2}\) 

⇒ θ = 30°, 150°. 

Since |sin θ| = \(\frac{3}{2}\) is > 1, the value sin θ = - \(\frac{3}{2}\) is inadmissible. 

∴ θ = 30°, 150°.

4 cos²θ = 3 

⇒ cos² θ= \(\frac{3}{4}\) 

⇒ cos θ= \(\pm \frac{\sqrt{3}}{2}\)

cos θ = \( \frac{\sqrt{3}}{2}\) ⇒ θ = 30°, 330° (∵ cos θ is +ve and so θ lies in 1st and 4th quad.) 

cos θ = \(- \frac{\sqrt{3}}{2}\)  ⇒ θ = 150°, 210° (∵ cos θ is –ve and so θ lies in 2nd and 3rd quad.) 

⇒ θ = 30°, 150°, 210°, 330°. 

Note: if the value of θ is α when θ lies in the Ist quadrant then it is 180° – α
, 180° + α and 360° – α, If θ lies in 2nd, 3rd, and 4th quadrant respectively.

tan 2A = cot (A – 12°)

or cot (90° – 2A) = cot (A – 12°)

On comparing

90° – 2A = A – 12 °

A + 2A = 90° + 12°

3A = 102°

A = 34°

The value of A is 34°

sec 4A = cosec (A – 15°)

or cosec (90° – 4A) = cosec (A – 15°)

On comparing

90° – 4A = A – 15°

A + 4A = 90° + 15°

5A = 105°

A = 21°

The value of A is 21°.

Dividing both the sides of the equation cos x − √3 sin x = 1 by

\(\sqrt{(1)^2 +(- \sqrt{3})^2}\) = 2 , we get

\(\frac{1}{2}\) cos x - \(\frac{\sqrt{3}}{2}\) sin x = \(\frac{1}{2}\) 

⇒ cos 60° cos x – sin 60° sin x = \(\frac{1}{2}\) 

⇒ cos (x + 60°) = cos 60° 

⇒ cos (x + 60°) = cos 60° = cos (360° – 60°) = cos (360° + 60°) 

⇒ x + 60° = 60° or 300° or 420° 

⇒ x = 0°, 240°, 360°.

Dividing both sides of the equation by

\(\sqrt{(\sqrt{3})^2 +(-1)^2}\) = \(\sqrt{4}\) = 2, we get \(\frac{\sqrt{3}}{2}\) sin θ - \(\frac{1}{2}\) cos θ = \(\frac{1}{\sqrt{2}}\)

⇒ cos 30° sin θ – sin 30° cos θ = \(\frac{1}{\sqrt{2}}\)

⇒ sin (θ – 30°) = sin \(\big(\) θ - \(\frac{π}{6}\) \(\big)\) = \(\frac{1}{\sqrt{2}}\) 

⇒ sin  \(\big(\) θ - \(\frac{π}{6}\) \(\big)\) = sin \(\frac{π}{4}\) 

⇒ θ - \(\frac{π}{6}\) = nπ +(-1)ⁿ \(\frac{π}{4}\) , n ∈ I 

⇒θ = nπ + \(\frac{π}{4}\) + (− 1)ⁿ \(\frac{π}{4}\) , n ∈ I .

Given, tan θ + sec θ = 4

⇒ \(\frac{sin\,\theta}{cos\,\theta}\)+ \(\frac{1}{cos\,\theta} = 4\) \(\frac{1+sin\,\theta}{cos\,\theta}=4\)

⇒ \(\frac{sin^2\frac{\theta}{2}+cos^2\frac{\theta}{2}+2sin\frac{\theta}{2}cos\frac{\theta}{2}}{(cos^2\frac{\theta}{2}-sin^2\frac{\theta}{2})}=4\)

(Using the formulas sin² θ + cos² θ = 1, sin 2θ = 2sin θ cos θ, cos 2θ = cos² – sin²θ)

⇒ \(\frac{\bigg(sin\frac{\theta}{2}+cos\frac{\theta}{2}\bigg)^2}{\bigg(cos\frac{\theta}{2}+sin\frac{\theta}{2}\bigg)\bigg(cos\frac{\theta}{2}-sin\frac{\theta}{2}\bigg)}=4\)

⇒\(\frac{\bigg(sin\frac{\theta}{2}+cos\frac{\theta}{2}\bigg)}{cos\frac{\theta}{2}-sin\frac{\theta}{2}}=4\) ⇒ \(\frac{1+tan\,\frac{\theta}{2}}{1-tan\,\frac{\theta}{2}}=4\)

⇒ 1 + tan\(\frac{\theta}{2}\) = 4 – 4 tan\(\frac{\theta}{2}\) ⇒ 5 tan\(\frac{\theta}{2}\) = 3 ⇒ tan\(\frac{\theta}{2}\) = \(\frac35\)

∴ sin θ = \(\frac{2\,tan\,\frac{\theta}{2}}{1+tan^2\,\frac{\theta}{2}}\) = \(\frac{2\times\frac35}{1+\frac{9}{25}}\) = \(\frac{\frac65}{\frac{34}{25}}\) = \(\frac{30}{34}\) = \(\frac{15}{17}.\)

Alternatively, 

Given, tan θ + sec θ = 4                ...(i) 

sec²  = 1 + tan² θ 

⇒ sec² θ – tan² q = 1                 ...(ii) 

eq. (ii) ÷ eq. (i) 

⇒ sec θ – tan θ = \(\frac14\)         ...(iii) 

eq. (i) + eq. (iii) 

⇒ 2 sec θ = \(\frac{17}{4}\) ⇒ sec θ = \(\frac{17}{8}\) 

⇒ cos θ = \(\frac{17}{8}\)

Now, use sin θ = \(\sqrt{1-cos^2\,\theta}\)

cos 3x + cos 2x = sin\(\frac{3}{2}\)x + sin \(\frac{1}{2}\)x 

⇒ 2 cos \(\frac{5}{2}\)x cos \(\frac{x}{2}\) = 2 sin x cos\(\frac{x}{2}\)

⇒ cos\(\frac{x}{2}\) \(\big[\) cos\(\frac{5x}{2}\) - sin x \(\big]\) = 0 

⇒ cos \(\frac{x}{2}\) = 0 or cos\(\frac{5x}{2}\) - sin x = 0

Now cos\(\frac{5x}{2}\) = 0 ⇒ \(\frac{x}{2}\) = \(\frac{\pi}{2}\) ⇒ x = π

and cos\(\frac{5x}{2}\) - sin x = 0

⇒ cos \(\frac{5x}{2}\) = sin x 

⇒ cos \(\frac{5x}{2}\) = cos( \(\frac{\pi}{2}\) - x )  or sin (2π + \(\frac{\pi}{2}\) -x)

⇒ \(\frac{5x}{2}\) = \(\frac{\pi}{2}\) - x or, \(\frac{5x}{2}\) = 2π +  \(\frac{\pi}{2}\) - x 

⇒\(\frac{7}{2}\)x = \(\frac{π}{2}\)  or \(\frac{7x}{2}\) = \(\frac{5π}{2}\) 

⇒ x = \(\frac{\pi}{7}\) or \(\frac{5\pi}{7}\)

∴ x = \(\frac{\pi}{7}\) , \(\frac{5\pi}{7}\) or π

tan 3x = tan \(\big(\) \(\frac{\pi}{2}\) - 5x \(\big)\) 

⇒ 3x = nπ + \(\big(\) \(\frac{\pi}{2}\) - 5x \(\big)\)  (∵ tan θ = tan α ⇒ θ = nπ + α)

⇒ 8x = (2n + 1)\(\frac{\pi}{2}\)  ⇒ x = (2n + 1) \(\frac{\pi}{16}\) 

∴  Putting n = 0, 1, 2, ..... 15, we see that the values of x between 0 and 2π are

x = \(\frac{\pi}{16}\) ,\(\frac{3\pi}{16}\) , \(\frac{5\pi}{16}\) ,..., \(\frac{31\pi}{16}\).

2sin² θ + 3sin θ = 0 

∴ sin θ (2sin θ + 3) = 0 

∴ sin θ = 0 or sin θ = \(\frac {-3}{2}\)

Since – 1 ≤ sin θ ≤ 1, 

sin θ = 0

\(\sqrt{1-\cos^2 \theta} =0\)...[∴ sin² θ = 1-cos²θ]

∴ 1- cos²θ = 0

∴ cos²θ = 1

∴ cos θ = ±1 …[∵ – 1 ≤ cos θ ≤ 1]

Given :

|sin x| = cos x 

⇒ sin² x = cos²x 

⇒ 1 – cos² x = cos² x 

⇒ 2 cos² x = 1 

⇒ cos x = + \(\frac{1}{\sqrt{2}}\) [∵ cos x = |sin x|, it cannot be negative] 

⇒ cos x = cos \(\frac{\pi}{4}\) 

⇒ x = 2nπ ± \(\frac{\pi}{4}\)

tan x + sec x = 2 cos x

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

⇒ 1 + sin x = 2 cos² x 

⇒ 1 + sin x = 2 (1 – sin² x) 

= 2 – 2 sin² x 

⇒ 2 sin² x + sin x – 1 = 0 

⇒ (sin x + 2) (2 sin x – 1) = 0 

⇒ (sin x + 2) = 0 or 2 sin x = 1 ⇒ sin x = –2 or sin x = \(\frac{1}{2}\) 

Since sin x = – 2 is inadmissible, therefore, sin x = \(\frac{1}{2}\) 

⇒ x = 30°, 150°, i.e. x = \(\frac{π}{6}\)  ,\(\frac{5π}{6}\).

∴ The number of solutions x ∈[0, π] are 2.

 sin 9θ = sin θ 

⇒ 9θ = 2nπ + θ or 9θ = (2n + 1) π - θ, n∈I (∵ sin θ = sin α ⇒ θ = nπ + (–1)ⁿ α, where n∈I)

⇒ θ = \(\frac{2n\pi}{8}\) or θ = \(\frac{(2n+1)\pi}{10}\) 

⇒ θ = \(\frac{n\pi}{4}\) or \(\frac{(2n+1)\pi}{10}\)

Correct option is (b) π/2 

3 sin 2A – 2sin 2B = 0 

sin 2B = 3/2 sin 2A …….(i) 

3 sin² A + 2 sin² B = 1 

3 sin² A = 1 – 2 sin² B 

3 sin² A = cos 2B ……(ii) 

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

= cos A (3 sin² A) – sin A (3/2 sin 2A) …..[From (i) and (ii)] 

= 3 cos A sin² A – 3/2 (sin A) (2 sin A cos A) 

= 3 cos A sin² A – 3 sin² A cos A 

= 0

= cos π/2

∴ A + 2B = π/2 ……..[∵ 0 < A + 2B < 3π/2]

Correct answer is (D) 1

Correct answer is (B) 3/4

Correct answer is (A) √3

Correct answer is (B) √2

Solution :

sin θ + cos θ = √3 (Given)

or (sin θ + cos θ)² = 3

or sin² θ + cos²θ + 2sinθ cosθ = 3

2sinθ cosθ = 2 [sin²θ + cos²θ = 1]

or sin θ cos θ = 1 = sin²θ + cos²θ

or 1 = (sin²θ + cos²θ)/sin θ cos θ

Therefore, tanθ + cotθ = 1

It is an imaginary line drawn from the eye of the observer to the point of the object viewed by the observer.

It is an imaginary line drawn from the eye of the observer to the point of the object viewed by the observer.

Correct option is: B) \(\frac{25}{7}\)

Sin A = \(\frac {24}{25}\)   

\(\therefore\) cos A = \(\sqrt {1-sin^2A}\)

= \(\sqrt{1-(\frac {24}{25})^2}\)

= \(\sqrt {\frac {25^2 - 24^2}{25^2}}\)

= \(\sqrt {\frac {625 - 576}{25}}\)

= \(\sqrt {\frac {49}{25}} = \frac 7{25}\)

\(\therefore\) Sec A = \(\frac 1{Cos\,A} = \frac {1} {\frac 7{25}} = \frac {25}7\)

Correct option is: B) \(\frac{25}{7}\)

Correct option is: B) G.P.

Sin \(30^\circ\) = \(\frac 12\)

sin \(90^\circ\) = 1

and sec \(60^\circ\) = 2

\(\because\) \(\frac 12\), 1 & 2 are in G.P

\(\therefore\) sin \(30^\circ\), sin \(90^\circ\) & sec \(60^\circ\) are in G.P

Correct option is: B) G.P.

LHS = sin (A + B) sin (A – B) 

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

= sin² A cos² B – cos² A sin² B 

= (1 – cos² A) cos² B – cos² A (1 – cos² B) 

= cos² B – cos² A cos² B – cos² A + cos² A cos² B 

= cos² B – cos² A = RHS

(i) tan(-225°) = -(tan 225°) 

= -(tan(180° + 45°)) 

= – tan 45° = – 1 

cot(-405°) = -(cot 405°) 

= – cot(360° + 45°) [∵ For 360° + 45° no change in T-ratio.] 

= -cot 45° 

= -1 

tan(-765°) = -tan 765° 

= -tan(2 x 360° + 45°) 

= -tan 45° 

= -1 

cot 675° = cot (360°+ 315°) 

= cot 315° 

= cot(360° – 45°) 

= -cot 45° 

= -1 

LHS = tan(-225°) cot(-405°) – tan(-765°) cot(675°) 

= (-1) (-1) – (-1) (-1) 

= 1 – 1

= 0 = RHS. 

Hence proved.

(ii) 2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7\pi}{6}\) cos² \(\frac{\pi}{3}\)= \(\frac{3}{2}\)

LHS = 2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7\pi}{6}\) cos² \(\frac{\pi}{3}\)= \(\frac{3}{2}\)

[∵ \(\frac{7\pi}{6}\)= 210°, 210° = 180° + 30°. For 180° + 30° no change in T-ratio. 210° lies in 3rd quadrant, cosec θ is negative.] 

= 2(sin\(\frac{\pi}{6}\))² + (cosec (180° + 30°))² (cos\(\frac{\pi}{3}\))²

= 2(\(\frac{1}{2}\))² + (-cosec 30°)².(\(\frac{1}{2}\))² 

= 2 x \(\frac{1}{4}+(-2)^2\frac{1}{4}\) 

= \(\frac{2}{4}+\frac{4}{4}=\frac{6}{4}\)

= \(\frac{6}{4}\)

= \(\frac{3}{2}\)

= RHS

(iii) sec(\(\frac{3\pi}{2}-\theta\)) = sec (270° – θ) = -cosec θ 

[∵ For 270° – θ change T-ratio. So add ‘co’ infront ‘sec’, it becomes ‘cosec’] 

sec(θ – \(\frac{5\pi}{2}\)) = sec(-(\(\frac{5\pi}{2}\) - θ))

= sec(\(\frac{5\pi}{2}\) – θ) [∵ sec(-θ) = θ] 

= sec(450° – θ) 

= sec (360° + (90° – θ))

= sec (90° – θ) 

= cosec θ 

[∵ For 90° – θ change in T-ratio. So add ‘co’ in front of ‘sec’ it becomes ‘cosec’] 

tan(\(\frac{5\pi}{2}\) + θ) = tan(450° + θ) 

[∵ For 90° + θ, change in T-ratio. So add ‘co’ in front of ‘tan’ it becomes ‘cot’] 

= tan (360° + (90° + θ)) 

= tan (90° + θ) 

= -cot θ

 tan(\(\theta-\frac{5\pi}{2}\)) = tan(-(\(\frac{5\pi}{2}\) - θ))

= -  tan (\(\frac{5\pi}{2}\) - θ) [∵ tan(-θ) = -tan θ] 

= -tan(450° – θ) 

= -tan(360° + (90° – θ)) 

= -tan(90° – θ) 

= -cot θ 

LHS = sec(\(\frac{3\pi}{2}-\theta\)) sec(\(\theta-\frac{5\pi}{2}\)) + tan(\(\frac{5\pi}{2}+\theta\)) tan(\(\theta-\frac{5\pi}{2}\))

= -cosec θ (cosec θ) + (-cot θ) (-cot θ) 

= -cosec² θ + cot² θ 

= -(1 + cot² θ) + cot² θ [∵ 1 + cot² θ = cosec² θ] 

= -1 

= RHS