SN Dey Class 11 Trigonometric Ratios of Multiple Angles MCQ

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Multiple Angles MCQ. Trigonometric Ratios MCQ

Choose the correct option:

Ex 1:  1-cos 2θ

(A) 2sin2θ  (B) 2cos2θ   (C) sin 2θ    (D) 2cos 2θ

Solution:

From the cos 2θ identity, we have

cos 2θ = 1-2sin2θ

⇒ 1-cos 2θ = 1- (1-2sin2θ) = 1-1+2sin2θ = 2sin2θ

∴ Option (A) is correct.

 

Ex 2:  $\dfrac{1+\cos 2\theta}{1-\cos2\theta}$

(A) sin2θ  (B) cos2θ   (C) tan2θ    (D) cot2θ

Solution:

Lets apply the formula cos 2θ = 2cos2θ -1 = 1- 2sin2θ. So we have

1+cos 2θ = 2cos2θ

1-cos 2θ = 2sin2θ

So $\dfrac{1+\cos 2\theta}{1-\cos2\theta}$ $=\dfrac{2\cos^2 \theta}{2\sin^2 \theta}$ $=\cot^2 \theta$.

∴ Option (D) is correct.

 

Ex 3:  $\dfrac{1-\cos 2\theta}{\sin 2\theta}$

(A) sin θ  (B) cos θ   (C) tan θ    (D) None of these

Solution:

We know that cos 2θ = 1 – 2sin2θ and sin 2θ = 2sinθ cosθ.

∴ $\dfrac{1-\cos 2\theta}{\sin 2\theta}$ $=\dfrac{2 \sin^2 \theta}{2\sin \theta \cos \theta}$ $=\dfrac{\sin \theta}{\cos \theta}$ = tan θ.

Thus, Option (C) is correct.

 

Ex 4:  $\dfrac{1+\tan^2 \theta}{1-\tan^2 \theta}$

(A) cos2θ  (B) cos 2θ   (C) sec2θ    (D) sec 2θ

Solution:

We know that $\cos 2 \theta =\dfrac{1-\tan^2 \theta}{1+\tan^2 \theta}$. This imples that

$\dfrac{1+\tan^2 \theta}{1-\tan^2 \theta}$ $=\dfrac{1}{\cos 2\theta}$ $=\sec 2\theta$.

∴ Option (D) is correct.

 

Ex 5:  cot 2A + tan A=

(A) sin 2A  (B) cosec 2A   (C) cos 2A    (D) sec 2A

Solution:

 

Also Read:

SN Dey Class 11 Set Theory MCQ Solutions

 

SN Dey Class 11 Limits MCQ Solutions

Ex 6:  State which of the following statement is not true?

(A) 4cos3A  = 3cos3A + cosA

(B) The formula of cos3A can be deduced from the formula sin3A=3sinA -4sin3A replacing A by (90°-A).

(C) cos 2θ = 2sin(π/4+θ) cos((π/4+θ)

(D) The formula of sin 2θ can be obtained replacing θ by (π/4+θ)  in the formula cos 2θ =2cos2θ-1.

Solution:

 

Ex 7:  If 2cosθ = x+1/x, state which of the following is the value of cos2θ?

(A) $x^2+\frac{1}{x^2}$

(B) $\frac{1}{2}(x^2+\frac{1}{x^2})$

(C) $\frac{1}{2}(x^3+\frac{1}{x^3})$

(D) $x^3+\frac{1}{x^3}$

Solution:

 

Ex 8:  If sinθ = $\dfrac{3}{5}$, state which of the following is the value of cos2θ?

(A) $\dfrac{7}{15}$  (B) $\dfrac{8}{25}$   (C) $\dfrac{2}{5}$    (D) $\dfrac{7}{25}$

Solution:

cos 2θ = 1-2sin2θ

= 1- 2 ×$\left( \dfrac{3}{5} \right)^2$

= 1- 2 ×$\dfrac{9}{25}$

= 1- $\dfrac{18}{25}$ $=\dfrac{25-18}{25}$ $=\dfrac{7}{25}$.

∴ Option (D) is correct.

 

Ex 9:  If tan A = 3, state which of the following is the value of tan 2A?

(A) $-\dfrac{3}{4}$  (B) $-\dfrac{4}{3}$   (C) $\dfrac{3}{4}$    (D) $\dfrac{4}{3}$

Solution:

tan 2A = $\dfrac{2\tan A}{1-\tan^2 A}$

$=\dfrac{2 \times 3}{1-9}$ as tanA = 3.

$=\dfrac{6}{-8}=-\dfrac{3}{4}$.

∴ Option (A) is correct.

 

Ex 10:  If sin A= $\dfrac{3}{5}$, state which of the following is the value of sin 3A?

(A) $\dfrac{17}{25}$  (B) $\dfrac{117}{125}$   (C) $\dfrac{24}{25}$    (D) $\dfrac{119}{125}$

Solution:

The sin 3A formula in terms of A is given below:

sin 3A =3sin A –4sin3A

= 3 ×$\dfrac{3}{5} – 4 \left( \dfrac{3}{5}\right)^3$

= $\dfrac{9}{5} – 4 \times \dfrac{27}{125}$

= $\dfrac{9}{5} – \dfrac{108}{125}$ $=\dfrac{225-108}{125}$ $=\dfrac{117}{125}$.

∴ Option (B) is correct.

 

Ex 11:  If cos A= $\dfrac{\sqrt{3}}{2}$, state which of the following is the value of cos 3A?

(A) $\dfrac{3\sqrt{3}}{8}$  (B) $-\dfrac{3\sqrt{3}}{8}$   (C) $0$    (D) $-1$

Solution:

The cos 3A formula in terms of A is given below:

cos 3A =4cos3A –3cos A

= $4 \left( \dfrac{\sqrt{3}}{2}\right)^3$ – 3 ×$\dfrac{\sqrt{3}}{2}$ 

= $4 \times \dfrac{3\sqrt{3}}{8}-\dfrac{3\sqrt{3}}{2}$

= $\dfrac{12\sqrt{3}}{8} – \dfrac{3\sqrt{3}}{2}$ $=\dfrac{12\sqrt{3}-12\sqrt{3}}{125}$ $=0$.

∴ Option (C) is correct.

 

Ex 12:  If tan x= $\dfrac{b}{a}$, state which of the following is the value of (a2+b2)sin2x?

(A) $ab$  (B) $2ab$   (C) $\dfrac{a}{b}$    (D) $\dfrac{2a}{b}$

Solution:

(a2+b2)sin 2x = (a2+b2) $\dfrac{2\tan x}{1+\tan^2 x}$

= (a2+b2) ×$\dfrac{2\frac{b}{a}}{1+\frac{b^2}{a^2}}$

= (a2+b2) ×$\dfrac{2\frac{b}{a}}{\frac{a^2+b^2}{a^2}}$

= 2(a2+b2) ×$\dfrac{b}{a} \times \dfrac{a^2}{a^2+b^2}$

= 2ab

∴ Option (B) is correct.