Question

Given that $(1+\sqrt{1+x})\tan y=1+\sqrt{1-x}$. Then $\sin 4y$ is equal to

• A. $4x$
• B. $2x$
• C. $x$
• D. none of these

Answer 1

The correct option is C $x$

$(1+\sqrt{1+x})\tan y=1+\sqrt{1-x}$ ...(1)
Squaring equation (1), we get
$(x+2+2\sqrt{1+x}){\tan }^{2}y=-x+2+2\sqrt{1-x}\Rightarrow x(1+{\tan }^{2}y)=-2(1+\sqrt{1+x}){\tan }^{2}y+2(1+\sqrt{1-x})$
$\Rightarrow x(1+{\tan }^{2}y)=-2(1+\sqrt{1+x}){\tan }^{2}y+2(1+\sqrt{1+x})\tan y$ ...{ from 1}

$\Rightarrow \frac{x(1+{\tan }^{2}y)}{2\tan y}=(1+\sqrt{1+x})(1-\tan y)$
$\Rightarrow \frac{x}{\sin 2y}=(1+\sqrt{1+x})(1-\frac{1+\sqrt{1-x}}{1+\sqrt{1+x}})$ ...{ from 1}

$\Rightarrow \sin 2y=\frac{x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\Rightarrow {\sin }^{2}2y=\frac{1}{4}(1+x+1-x+2\sqrt{1-x^2})\Rightarrow \frac{1+\cos 4y}{2}=\frac{1+\sqrt{1-x^2}}{2}\Rightarrow \cos 4y=\sqrt{1-x^2}\Rightarrow \sin 4y=x$

Ans: C