Question

Given that $t+\sqrt{(1+x)}\tan x=1+\sqrt{(1-x)}$. The value of $\sin 4x$ is equal to

• A. $x$
• B. $2x$
• C. $3x$
• D. $4x$

Answer 1

Answer: (A)

$x$

$⟹tany=\frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}$

Let $x=cos\theta$

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

$\sqrt{1+x}=\sqrt{2}cos\left(\frac{\theta }{2}\right)$

$\Rightarrow tany=\frac{\sqrt{2}|\frac{1}{\sqrt{2}}+sin\frac{\theta }{2}|}{\sqrt{2}|\frac{1}{\sqrt{2}}+cos\frac{\theta }{2}|}$

$=\frac{sin\frac{\pi }{4}+sin\frac{\theta }{2}}{cos\frac{\pi }{4}+cos\frac{\theta }{2}}$

$\Rightarrow tany=tan(\frac{\pi }{8}+\frac{\theta }{4})$

$\Rightarrow 4y=\frac{\pi }{2}+\theta$

$\Rightarrow sin4y=cos\theta =x$