SN Dey Class 11 Solutions | SN Dey Class 11 Trigonometry Solutions | SN Dey Class 11 Trigonometric Ratios of Multiple Angles Multiple Choice Type Questions | Class 11 Trigonometric Ratios of Multiple Angles MCQ | SN Dey Solutions | S N Dey Math Solutions | Trigonometric Ratios of Multiple Angles S N Dey Solutions | WBCHSE Class 11 Math Solutions | Trigonometric Ratios of Multiple Angles objective questions | SN Dey Trigonometry
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
|
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.
