ML Aggarwal Solution Class 10 Chapter 18 Trigonometric Identities MCQs

MCQs 

Choose the correct answer from the given four options (1 to 12) :

Question 1

$\cot ^{2} \theta-\frac{1}{\sin ^{2} \theta}$ is equal to

(a) 1

(b) -1

(c) $\sin ^{2} \theta$

(d) $\sec ^{2} \theta$

Sol :

$\cot ^{2} \theta-\frac{1}{\sin ^{2} \theta}$

$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-\frac{1}{\sin ^{2} \theta}$

$\frac{\cos ^{2} \theta-1}{\sin ^{2} \theta}=\frac{-\sin ^{2} \theta}{\sin ^{2} \theta}=-1$

Ans (b)

Question 2

$\left(\sec ^{2} \theta-1\right)\left(1-\operatorname{cosec}^{2} \theta\right)$ is equal to

(a) – 1

(b) 1

(c) 0

(d) 2

Sol :

$\left(\sec ^{2} \theta-1\right)\left(1-\operatorname{cosec}^{2} \theta\right)$

$=\left(\frac{1}{\cos ^{2} \theta}-1\right)\left(1-\frac{1}{\sin ^{2} \theta}\right)$

$=\frac{1-\cos ^{2} \theta}{\cos ^{2} \theta} \times \frac{\sin ^{2} \theta-1}{\sin ^{2} \theta}$

$=\frac{-\sin ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}=-1$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

Ans (a)

Question 3

$\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}$ is equal to

(a) $2 \sin ^{2} \theta$

(b) $2 \cos ^{2} \theta$

(c) $\sin ^{2} \theta$

(d) $\cos ^{2} \theta$

Sol :

$\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}$

$\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos ^{2} \theta}}$

$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta+\cos ^{2} \theta}$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

$=\frac{\sin ^{2} \theta}{1}=\sin ^{2} \theta$

Ans (c)

Question 4

$(\cos \theta+\sin \theta)^{2}+(\cos \theta-\sin \theta)^{2}$ is equal to

(a) – 2

(b) 0

(c) 1

(d) 2

Sol :

$(\cos \theta+\sin \theta)^{2}+(\cos \theta-\sin \theta)^{2}$

$=\cos ^{2} \theta+\sin ^{2} \theta+2 \sin \theta \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta-2 \sin \theta \cos \theta$

$=2\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$

$=2 \times 1=2(d)$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

Question 5

(sec A + tan A) (1 – sin A) is equal to

(a) sec A

(b) sin A

(c) cosec A

(d) cos A

Sol :

(sec A + tan A) (1 – sin A)

$=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$

$=\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}} \times 1-\sin \mathrm{A}$

$=\frac{(1+\sin \mathrm{A})(1-\sin \mathrm{A})}{\cos \mathrm{A}}$

$=\frac{1-\sin ^{2} A}{\cos A}=\frac{\cos ^{2} A}{\cos A}=\cos A$

Ans (d)

Question 6

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$ is equal to

(a) $\sec ^{2} A$

(b) -1

(c) $\cot ^{2} A$

(d) $\tan ^{2} A$

Sol :

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}$

$=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}$

$=\frac{1}{\cos ^{2} A} \times \frac{\sin ^{2} A}{1}$

$=\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A$

Ans (d)

Question 7

If sec θ – tan θ = k, then the value of sec θ + tan θ is

(a) $1-\frac{1}{k}$

(b) $1-k$

(c) $1+k$

(d) $\frac{1}{k}$

Sol :

sec θ – tan θ = k

$\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}=k$

$\frac{1-\sin \theta}{\cos \theta}=k$

Squaring both sides, we get

$\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}=(k)^{2} \Rightarrow \frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}=k^{2}$

$=\frac{(1-\sin \theta)^{2}}{1-\sin ^{2} \theta}=k^{2}$

$\Rightarrow \frac{(1-\sin \theta)^{2}}{(1+\sin \theta)(1-\sin \theta)}=k^{2}$

$=\frac{1-\sin \theta}{1+\sin \theta}=k^{2}$

$\Rightarrow \frac{1+\sin \theta}{1-\sin \theta}=\frac{1}{k^{2}}$

$\Rightarrow \sqrt{\frac{(1+\sin \theta)}{1-\sin \theta}}=\frac{1}{k}$

Multiplying and dividing by $(1+\sin \theta)$

$=\sqrt{\left[\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta}\right]}=\frac{1}{k}$

$\Rightarrow \sqrt{\left[\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}\right]}=\frac{1}{k}$

$\frac{1+\sin \theta}{\cos \theta}=\frac{1}{k}=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=\frac{1}{k}$

$\Rightarrow \sec \theta+\tan \theta=\frac{1}{k}$

Ans (d)

Question 8

Which of the following is true for all values of θ (0° < θ < 90°):

(a) $\cos ^{2} \theta-\sin ^{2} \theta=1$

(b) $\operatorname{cosec}^{2} \theta-\sec ^{2} \theta=1$

(c) $\sec ^{2} \theta-\tan ^{2} \theta=1$

(c) $\cot ^{2} \theta-\tan ^{2} \theta=1$

Sol :

$\therefore \sec ^{2} \theta-\tan ^{2} \theta=1$ is true for all values of θ as it is an identity.

(0° < θ < 90°)

Ans (c)

Question 9

If θ is an acute angle of a right triangle, then the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is

(a) 0

(b) 2 sin θ cos θ

(c) 1

(c) $2 \sin ^{2} \theta$

Sol :

sin θ cos (90° – θ) + cos θ sin (90° – θ)

= sin θ sin θ + cos θ cos θ

{ ∵ sin(90° – θ) = cosθ, cos (90° – θ) = sin θ }

$=\sin ^{2} \theta+\cos ^{2} \theta=1$

Ans (c)

Question 10

The value of cos 65° sin 25° + sin 65° cos 25° is

(a) 0

(b) 1

(b) 2

(d) 4

Sol :

cos 65° sin 25° + sin 65° cos 25°

= cos (90° – 25°) sin 25° + sin (90° – 25°) cos 25°

= sin 25° . sin 25° + cos 25° . cos 25°

$=\sin ^{2} 25^{\circ}+\cos ^{2} 25^{\circ}$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

=1

Ans (b)

Question 11

The value of $3 \tan ^{2} 26^{\circ}-3 \operatorname{cosec}^{2} 64^{\circ}$ is

(a) 0

(b) 3

(c) – 3

(d) – 1

Sol :

$3 \tan ^{2} 26^{\circ}-3 \operatorname{cosec}^{2} 64^{\circ}$

$=3 \tan ^{2} 26^{\circ}-3 \operatorname{cosec}\left(90^{\circ}-26^{\circ}\right)$

$=3 \tan ^{2} 26^{\circ}-3 \sec ^{2} 26^{\circ}$

$=3\left(\tan ^{2} 26^{\circ}-\sec ^{2} 26^{\circ}\right)$

$=3 \times(-1)=-3$  $\left\{\because \sec ^{2} \theta-\tan ^{2} \theta=1\right\}$

Ans (c)

Question 12

The value of $\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1$ is

(a) $-\cot \theta$

(b) $-\sin ^{2} \theta$

(c) $-\cos ^{2} \theta$

(d) $-\operatorname{cosec}^{2} \theta$

Sol :

$\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1$

$=\frac{\cos \theta \sin \theta}{\frac{\sin \theta}{\cos \theta}}-1$

$=\frac{\sin \theta \cos \theta \times \cos \theta}{\sin \theta}-1$

$=\cos ^{2} \theta-1=-\left(1-\cos ^{2} \theta\right)$

$=-\sin ^{2} \theta$

Ans (b)