If 1+sqrt{1+a}=(1+sqrt{1-a}) cot alpha and 0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation: $(1+\sqrt{1+a})=(1+\sqrt{1-a}) \cot \alpha$ Rewrite as: $(1+\sqrt{1+a}) \sin \alpha = (1+\sqrt{1-a}) \cos \alpha$ Rearranging

Vedclass · Accepted Answer

$a$

Vedclass · Answer

(A) Given the equation: $(1+\sqrt{1+a})=(1+\sqrt{1-a}) \cot \alpha$ Rewrite as: $(1+\sqrt{1+a}) \sin \alpha = (1+\sqrt{1-a}) \cos \alpha$ Rearranging terms: $(\sin \alpha - \cos \alpha) = \sqrt{1-a} \cos \alpha - \sqrt{1+a} \sin \alpha$ Squaring both sides: $(\sin \alpha - \cos \alpha)^2 = (\sqrt{1-a} \cos \alpha - \sqrt{1+a} \sin \alpha)^2$ $1 - \sin 2 \alpha = (1-a) \cos^2 \alpha + (1+a) \sin^2 \alpha - 2\sqrt{1-a^2} \sin \alpha \cos \alpha$ $1 - \sin 2 \alpha = (\cos^2 \alpha + \sin^2 \alpha) + a(\sin^2 \alpha - \cos^2 \alpha) - \sqrt{1-a^2} \sin 2 \alpha$ $1 - \sin 2 \alpha = 1 - a \cos 2 \alpha - \sqrt{1-a^2} \sin 2 \alpha$ $a \cos 2 \alpha - \sin 2 \alpha = -\sqrt{1-a^2} \sin 2 \alpha$ Squaring again: $a^2 \cos^2 2 \alpha + \sin^2 2 \alpha - 2a \sin 2 \alpha \cos 2 \alpha = (1-a^2) \sin^2 2 \alpha$ $a^2 \cos^2 2 \alpha + \sin^2 2 \alpha - a \sin 4 \alpha = \sin^2 2 \alpha - a^2 \sin^2 2 \alpha$ $a^2 \cos^2 2 \alpha + a^2 \sin^2 2 \alpha = a \sin 4 \alpha$ $a^2(1) = a \sin 4 \alpha$ Therefore,$\sin 4 \alpha = a$.

If $1+\sqrt{1+a}=(1+\sqrt{1-a}) \cot \alpha$ and $0 < a < 1$,then $\sin 4 \alpha=$

Similar Questions

If a regular pentagon and a regular decagon have the same perimeter,then the ratio of their areas is

If $(\sin \theta - \operatorname{cosec} \theta)^2 + (\cos \theta + \sec \theta)^2 = 5$ and $\theta$ lies in the third quadrant,then $(\sin \theta + \cos \theta)^3 = $

Which of the following statements is/are correct for $0 < \theta < \frac{\pi}{2}$?

If $\cos x + \cos y + \cos \alpha = 0$ and $\sin x + \sin y + \sin \alpha = 0,$ then $\cot \left( \frac{x + y}{2} \right) = $

Let $P = \{ \theta : \sin \theta - \cos \theta = \sqrt{2} \cos \theta \}$ and $Q = \{ \theta : \sin \theta + \cos \theta = \sqrt{2} \sin \theta \}$ be two sets. Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module