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Given sin 52° = a

We know that sin θ = cos (90° - θ)

Here, θ = 52°

⇒ cos (90° - 52°) = a

⇒ cos 38° = a

Given cos 47° = a

We know that cos θ = sin (90° - θ)

Here, θ = 47°

⇒ sin (90° - 47°) = a

⇒ sin 43° = a

Answer : (a) nπ 

4 sin⁴ x + cos⁴ x = 1 

⇒ 4 sin⁴ x + (1 – sin²x) ² = 1 

⇒ 4 sin⁴x + 1 + sin⁴x – 2 sin²x = 1 

⇒ 5 sin⁴x – 2 sin²x = 0 

⇒ sin²x (5 sin²x – 2) = 0 

⇒ sin²x = 0 or 5 sin²x – 2 = 0

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

⇒ x = nπ or x = sin^–1 ( \(\sqrt{\frac{2}{5}}\) )

= nπ + (–1)ⁿ sin^–1  ( \(\sqrt{\frac{2}{5}}\) ), n ∈ I

∴ x = nπ is one of the solutions of the given equation.

Answer : (c) 6 

3 sin² x – 7 sin x + 2 = 0 

⇒ (3 sin x – 1) (sin x – 2) = 0 

⇒ (3 sin x – 1) = 0 or (sin x – 2) = 0 

∵ sin x = 2 is inadmissible, therefore, sin x = \(\frac{1}{3}\)

Since, sin x = sin α where sin α = \(\frac{1}{3}\) , so α lies in the 1st quadrant 

⇒ x = nπ + (–1)ⁿ α, n∈I, where 0 < α < π/2 

Since x lies in the interval [0, 5π], so we have one value of x corresponding to each of the values 0, 1, 2, 3, 4, 5 or n. 

∴ The number of values of x in the interval [0, 5π] is 6.

(d) 1

sin⁴x + sin²x = 1 

⇒ sin⁴x = 1 – sin²x 

⇒ sin⁴x = cos²x              ...(i) 

∴ cot⁴x + cot²x 

= cot²x (1 + cot²x) = cot²x . cosec²x

\(=\frac{cos^2\,x}{sin^2\,x.sin^2\,x}=\frac{cos^2\,x}{sin^4\,x}\)

= \(=\frac{cos^2\,x}{cos^2\,x}\) = 1.                       [Using (i)]

(c) \(±\sqrt{c^2+b^2-a^2}\)

Given, a sin θ – b cos θ = c 

(a cos θ – b sin θ)² 

= a² cos²θ – 2ab cos θ sin θ + b² sin²θ 

= a² (1 – sin²θ) – 2(a sin θ) (b cos θ) + b² (1 – cos²θ) 

= a² – a² sin²θ – 2 (a sin θ) (b cos θ) + b² – b² cos²θ 

= a² + b² – [a² sin²θ + 2 (a sin θ) (b cos θ) + b² cos²θ] 

= a² + b² – (a sin θ + b cos θ)² = a² + b² – c² 

⇒ (a cos θ – b sin θ) = \(±\sqrt{c^2+b^2-a^2}\)

(d) cosec² A – cot² A.

sin A cos A tan A + cos A sin A cot A 

= sin A cos A . \(\frac{sin\,A}{cos\,A}\) + cos A sin A \(\frac{cos\,A}{sin\,A}\)

= sin² A + cos² A = 1 = cosec² A – cot² A.

(a) 0

cos α sin (β – γ) + cos β sin (γ – α) + cos γ sin (α – β) 

= cos α [sin β cos γ – cos β sin γ] + cos β [sin γ cos α – cos α sin β] + cos γ [sin α cos β – cos α sin β] 

= cos α sin β cos γ – cos α cos β sin γ + cos β sin γ cos α – cos β cos α sin γ + cos γ sin α cosβ – cos γ cos α sin β 

= 0.

tan^-1[tan3π/4] = tan^-1(tan(π-π/4))

= tan^-1(tan π/4)

Since tan^-1(tanx) = x if x ∈ (-π/2,π/2)

Here π/4 ∈ (-π/2,π/2)

= π/4

(2) 90°

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

α + β = 90°

cos 3α + cos 3β + cos 3γ = (4 cos³α - 3 cos³α ) + (4 cos³ β – 3 cos β) + (4 cos³ γ – 3 cos γ) 

= 4(cos³ α + cos³ β + cos3 γ) – 3(cos α + cos β + cos γ) 

= 4(cos³ α + cos³ β + cos³ γ) – 3 × 0 = 4 (cos³ α + cos³ β + cos³ γ) 

= 4 × 3 cos α cos β cos γ       (∵ a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc) 

= 12 cos α cos β cos γ.

Consider a² + b² = sin² α + sin² β + 2 sin α sin β + cos² α + cos² β + 2 cos α cos β 

a² + b² = (sin² α + cos² α) + (sin² β + cos² β) + 2[cos α cos β + sin α sin β] 

a² + b² = 1 + 1 + 2 cos(α – β) 

∴ cos(α – β) = \(\frac{a^2+b^2-2}{2}\)

Correct option (a) 1

Explanation:

cotα.cot(2n–1)α = cotα.cot(2nα – α) = cotα.cot (π/2 -α) = cotα.tanα = 1

Product of terms equidistant from the beginning and end is 1 The middle term is cotnα = cotπ/4 = 1

Given, 

m² = a² cos² θ + 2 ab sin θ cos θ + b² sin²θ

and

n² = a² sin² θ-2 ab sin θ cos θ + b²cos² θ

Adding equations (i) and (ii)

m² + n² = a² (cos² θ + sin² θ) + b²(cos²θ + sin² θ)

= a² (1) + b² (1)

= a² + b² = RHS

Answer: (b)  2nπ ±  cos^–1 \(\big(\frac{1}{3}\big)\) 

3 sin² x + 10 cos x – 6 = 0 

⇒ 3 (1 – cos² x) + 10 cos x – 6 = 0 

⇒ – 3 cos² x + 10 cos x – 3 = 0 

⇒ (cos x – 3) (1 – 3 cos x) = 0 

⇒ cos x = 3 or cos x = \(\frac{1}{3}\)

Since cos x = 3 is inadmissible, therefore, cos x = \(\frac{1}{3}\)

⇒ x = cos^–1 \(\big(\frac{1}{3}\big)\) 

The general solution is  

x = 2nπ ±  cos^–1 \(\big(\frac{1}{3}\big)\) 

Answer: (b) = \(\frac{π}{3}\)

tan θ + sec θ = √3

⇒ \(\frac{sin\,\theta}{cos\,\theta}\) + \(\frac{1}{cos\,\theta}\) = √3 

⇒ sin θ + 1 = √3 cos θ

⇒ \(\frac{\sqrt{3}}{2}\) cos θ – \(\frac{1}{2}\) sin θ = \(\frac{1}{2}\) 

⇒ cos (θ + \(\frac{π}{6}\)) = cos\(\frac{π}{3}\)

⇒ Principal value of θ + \(\frac{π}{6}\) = \(\frac{π}{3}\).

Answer : (d) \(\big( \frac{5π}{4},\frac{7π}{4}\big)\)

cos² θ + sin θ + 1 = 0 

⇒ (1 – sin² θ) + sin θ + 1 = 0 

⇒ sin² θ – sin θ – 2 = 0 

⇒ (sin θ + 1) (sin θ – 2) = 0 

⇒ (sin θ + 1) = 0 or (sin θ – 2) = 0 

⇒ sin θ = –1 (∵ sin θ = 2 is inadmissible) 

⇒ sin θ = sin \(\frac{3π}{2}\) 

⇒ θ = \(\frac{3π}{2}\) ∈ \(\big( \frac{5π}{4},\frac{7π}{4}\big)\) 

cosec 780° = cosec (720° + 60°) 

= cosec (2 x 360° + 60°) 

= cosec 60°

= 2√3

sec (-855°) = sec (855°) 

= sec (720°+135°)

= sec (2 x360°+ 135°) = sec 135° 

= sec (90° + 45°) 

= – cosec 45°

= -√2

sec 240° = sec (180° + 60°) 

= – sec 60° 

= – 2

Correct option is: C) 1

<A = \(75^\circ\) & B = \(30^\circ\)

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

\(\therefore\) tan (A-B) = tan (\(45^\circ\)) = 1

Correct option is: C) 1

Correct option is: B) \(\frac{7}{5}\)

We have 

sin x = \(\frac 57\)

\(\therefore\) cosec x = \(\frac 1{sin \, x} = \frac {1}{\frac 57} = \frac 75\)

Correct option is: B) \(\frac{7}{5}\)

Correct option is: D) 90°

\(\because\) tan \(\theta\) is not defined at \(90^\circ\).

\(\therefore\) Among given angles tan \(\theta\) is not defined at only \(90^\circ\).

Correct option is: D) 90°

Correct option is: A) \(\frac {16}4\)

We have tan \(\theta\) = \(\frac 1{\sqrt3} = tan \, 30^\circ\)

 = \(\theta\) = \(30^\circ\)

Now \(7\, sin^2\theta + 3\, cos^2\theta = 7\, sin^2\,30^\circ + 3\, cos^2\, 30^\circ\).

= 7 \(\times (\frac 12)^2 + 3 (\frac {\sqrt3}{2})^2 \) (\(\because\) sin \(30^\circ\) = \(\frac 12\) & cos \(30^\circ\) = \(\frac {\sqrt3}2\).

= \(\frac 74 + 3 \times \frac 34 = \frac 74 + \frac 94 = \frac {16}4 = 4\)

Alternative method :

We have tan \(\theta\) = \(\frac 1{\sqrt3} \)

\(\therefore\) \(Sec^2\theta = 1 + tan^2\theta = 1 + (\frac 1{\sqrt3})^2 = 1+ \frac 13 = \frac 43\)

Now, \(7\, sin^2\theta + 3\, cos^2\theta = \frac {7\, sin^2\theta + 3\, cos^2\theta}{cos^2\theta}.cos^2\theta\).

= \(\frac {(7\, tan^2\theta + 3)}{sec^2\theta}\) (\(\because\) \(\frac 1{cos\theta} = sec \theta\))

= \(\frac {7\times \frac 13 + 3}{\frac 43}\)

= \(\frac {\frac {7+9}{3}}{\frac 43} = \frac {16}4 = 4\).

Correct option is: A) \(\frac{16}{4}\)

Correct option is: A) 3

sec \(\theta\) + tan \(\theta\) = \(\frac{1}{3}\)

= ( sec \(\theta\) + tan \(\theta\)) ( sec \(\theta\) - tan \(\theta\)) = \(\frac{1}{3}\) ( sec \(\theta\) - tan \(\theta\))

(Multiplying both sides by (sec \(\theta\) - tan \(\theta\)))

= \(sec^2\theta - tan^2\theta = \frac 13 ( sec\, \theta - tan \, \theta)\) (\(\because\) (a + b ) (a - b) = \(a^2-b^2\))

= sec \(\theta\) - tan \(\theta\) = 3 (\(sec^2 \theta - tan^2\theta\)) = 3 (\(\because\) \(Sec^2 \theta - tan ^2 \theta = 1\))

Hence, if sec \(\theta\) + tan \(\theta\) = \(\frac{1}{3}\) then sec \(\theta\) - tan \(\theta\) = 3

Correct option is: A) 3

tan² θ + cot² θ = tan² θ + 1/tan² θ 

(tan θ)² + (1/tan θ)²

= (tan θ - 1/ tan θ)² + 2tan θ. 1/ tan θ

                ...[∴a² + b² = (a-b)² + 2ab]

= (tan θ - 1/ tan θ)² + 2 ≥ 2 for all θ ∈R.

We know that, sin² θ = 1 – cos θ sin² 36° = 1 – cos² 36°

= 1-(\(\frac{\sqrt{5+1}}{4}\))²

^{=\(\frac{16-(5+1+2\sqrt5)}{16}\)}

= ^{\(\frac{10-2\sqrt5}{16}\)}

∴ sin 36° = \(\frac{\sqrt{10-2\sqrt5}}{4}\)……[∵ sin 36° is positive]

L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A] 

= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A] Let(n+2)Aa and(n+l)Ab …(i) 

∴ L.H.S. = cos a. cos b + sin a. sin b 

= cos (a — b) 

= cos [(n + 2)A — (n + I )A] 

…[From (i)] 

cos[(n+2 – n – 1)A] 

= cos A 

= R.H.S.

L.H.S. = cos(x + y). cos(x – y) 

= (cos x cos y – sin x sin y). (cos x cos y + sin x sin y) 

= cos² x cos² y – sin² x sin² y 

…[∵ (a – b) (a + b) = a² – b² ] 

= (1 – sin² x) cos² y – sin² x (1 – cos² y) 

…[∵ sin² e + cos² 0 = 1] 

= cos² y – cos² y sin² x – sin² x + sin² x cos² y 

= cos² y – sin² x 

=R.H.S.

L.H.S. = cos(x + y). cos(x – y) 

= (cos x cos y – sin x sin y). (cos x cos y + sin x sin y) 

= cos² x cos² y – sin² x sin² y 

…[∵ (a – b) (a + b) = a² – b²] 

= (1 – sin² x) cos² y – sin² x (1 – cos² y) 

…[∵ sin² e + cos² 0 = 1] 

= cos² y – cos² y sin² x – sin² x + sin² x cos² y 

= cos² y – sin² x 

=R.H.S

Correct answer is (C) 1/2

Solution : We know that sin²θ + cos²θ = 1

Therefore, (sin²θ + cos²θ)³ = 1

or, (sin²θ)³ + (cos²θ)³ + 3sin²θ cos²θ (sin²θ + cos²θ) = 1

or, sin⁶θ + cos⁶θ + 3sin²θ cos²θ = 1

(i) sin A + sin 2A = 2 sin(\(\frac{A+2A}{2}\)) cos(\(\frac{A-2A}{2}\)) 

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

= 2 sin\(\frac{3A}{2}\) cos\(\frac{A}{2}\) [∵ cos(-θ) = cos θ]

(ii) cos 2A + cos 4A = 2 cos(\(\frac{2A+4A}{2}\)) cos(\(\frac{2A-4A}{2}\)) 

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

= 2 cos(\(\frac{6A}{2}\)) cos(\(\frac{6-2A}{2}\)) 

= 2 cos(3A) cos (-A) [∵ cos(-θ) = cos θ] 

= 2 cos 3A cos A

(iii) sin 6θ – sin 2θ = 2 cos(\(\frac{6\theta+2\theta}{2}\)) cos(\(\frac{6\theta-2\theta}{2}\)) 

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

= 2 cos(\(\frac{8\theta}{2}\)) sin(\(\frac{4\theta}{2}\)) 

= 2 cos 4θ sin 2θ

(iv) cos 2θ – cos θ = -2 sin(\(\frac{2\theta+\theta}{2}\)) sin(\(\frac{2\theta-\theta}{2}\)) 

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

= -2 sin(\(\frac{3\theta}{2}\)) sin(\(\frac{\theta}{2}\))

(c) 0

Given sin A + cos A = 1 

Squaring both sides we get 

sin² A + cos² A + 2 sin A cos A = 1

1 + sin 2A = 1 

sin 2A = 0

cos² 17° - sin² 73° = cos² 17° - sin²(90° -17°) 

= cos²17° - cos² 17° = 0

(a) -1

sec A (sin(270° + A)) = \(\frac{1}{cosA}\)(-cos A) = - 1

Correct option is : (B) 0

cos 1° cos 2° cos 3° … cos 179° 

= cos 1° cos 2° cos 3° … cos 90°… cos 179° 

= 0 …[∵ cos 90° = 0]

(d) \(\frac{1}{4}\)

sin 15° cos 15° = \(\frac{1}{2}\)(2 sin 15° cos 15°) 

= \(\frac{1}{2}\)(sin 30°) 

= \(\frac{1}{2}\)(\(\frac{1}{2}\))

= \(\frac{1}{4}\)

cos² x + cos² (x + 120°) + cos² (x – 120°) 

= \(\frac{1+cos2x}{2}+\frac{1+cos2(X+120°)}{2} + \frac{1+cos 2(x-120°)}{2}\)

...[∵cos²θ = \(\frac{1+cos \,2θ}{2}\)

\(\frac{3}{2}+\frac{1}{2}\)[cos 2x + cos(2x + 240°) + cos(2x 240°)]

= \(\frac{3}{2}+\frac{1}{2}\)(cos 2x + cos 2x cos 240°— sin 2x sin 240° + cos 2x cos 240° + sin 2x sin 240°)

=\(\frac{3}{2}+\frac{1}{2}\)(cos 2x + 2 cos 2x cos 240°)

=\(\frac{3}{2}+\frac{1}{2}\)[cos 2x + 2 cos 2x cos( 180° + 60°)]

= \(\frac{3}{2}+\frac{1}{2}\)[cos 2x + 2cos 2x(-cos 600)]

= \(\frac{3}{2}+\frac{1}{2}\)[cos 2x —2 cos 2x(\(\frac{1}{2}\))]

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

= \(\frac{3}{2}+\frac{1}{2}\)(0)

= \(\frac{3}{2}\)= R.H.S

(b) 0

∵ cos 90° = 0 

∴ cos 1°. cot 2°. ....... . cos 90° . cos 91° ..... . cos 100° = 0.

(a) \(\frac{1}{\sqrt{2}}\)

sin 28° cos 17° + cos 28° sin 17° = sin(28° + 17°) 

This is of the form sin(A + B), A = 28°, B = 17° 

= sin 45°

= \(\frac{1}{\sqrt{2}}\)

(b) \(-\frac{1}{2}\)

4 cos³ 40° – 3 cos 40° 

= cos (3 x 40°) [∵ cos 3A = 4 cos³ A – 3 cos A] 

= cos 120° 

= cos (180° – 60°) 

= -cos 60° 

= - \(\frac{1}{2}\)

(a) 57° 16′ 22′′

π radians = 180°

⇒ 1 radian = \(\frac{180}{π} \) degree = \(\frac{180}{22}\times7 \) degree

= \(\frac{630}{11}\)degree = \(57\frac{3}{11}\) degree

= 57° + \(\frac{3}{11}\times60\) min = 57° + \(\frac{180}{11}\) min

= 57° + 16\(\frac4{11}\)

= 57°+ 16' + \(\frac4{11}\) x 60 s

= 57° + 16′ + 21.8′′ 

= 57° 16′ 22′′ (approx).

(d) √3

We know that sin 2A = \(\frac{2tanA}{1+tan^2A}\)

\(\frac{2tan30°}{1+tan^230°}\) = sin(2 x 30°) = sin 60° = \(\frac{\sqrt{3}}{2}\)

= tan 2A 

= tan 60° 

= √3

(c) \(\frac{\pi}{3}\)

Let cosec^-1 (\(\frac{2}{\sqrt{3}}\)) 

\(\frac{2}{\sqrt{3}}\) = cosec A 

cosec A = \(\frac{2}{\sqrt{3}}\)

sin A = \(\frac{\sqrt{3}}{2}\) = sin 60° 

∴ A = 60° = \(\frac{\pi}{3}\)

(b) 0

Given sin A = \(\frac{1}{2}\)

sin A = sin 30°

∴ A = 30° 

[∵ 4 cos³ A – 3 cos A = cos 3A] 

= cos(3 x 30°) 

= cos 90° 

= 0