 QUESTION

# If the value of $(1 + \sqrt {1 + x} \tan y = 1 + \sqrt {1 - x} )$, then $\sin 4y$ is equal to A.4xB.2xC.xD.None of these

Hint: Use basic trigonometric conversions like $\sin C + \sin D = 2\sin (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2}),\cos C + \cos D = 2\cos (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$. Use this to find the value of $\sin 4y$.
$\tan y = \dfrac{{1 + \sqrt {1 - x} }}{{1 + \sqrt {1 + x} }}$……………..(i)

Let $x = \cos \theta$
$\Rightarrow $1 + x = 1 + \cos \theta = 2{\cos ^2}(\dfrac{\theta }{2}) \Rightarrow \sqrt {1 + x} = \sqrt 2 \cos \dfrac{\theta }{2} Similarly , \Rightarrow 1 - x = 1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2} \Rightarrow \sqrt {1 - x} = \sqrt 2 \sin \dfrac{\theta }{2}, Hence substituting the value of \sqrt {1 - x} in equation(i) We get, \tan y = \dfrac{{1 + \sqrt {1 - x} }}{{1 + \sqrt {1 + x} }} = \dfrac{{1 + \sqrt 2 \sin (\dfrac{\theta }{2})}}{{1 + \sqrt 2 \cos \dfrac{\theta }{2}}} \\ = \dfrac{{\sqrt 2 (\dfrac{1}{{\sqrt 2 }} + \sin (\dfrac{\theta }{2}))}}{{\sqrt 2 (\dfrac{1}{{\sqrt 2 }} + \cos (\dfrac{\theta }{2}))}} \\ As we know that \dfrac{1}{{\sqrt 2 }}can be written as \dfrac{\pi }{4}. Hence, \tan y = \dfrac{{\sin \dfrac{\pi }{4} + \sin \dfrac{\theta }{2}}}{{\cos \dfrac{\pi }{4} + \cos \dfrac{\theta }{2}}} We know that , \sin C + \sin D = 2\sin (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2}) \\ \cos C + \cos D = 2\cos (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2}) \\ Hence substituting the value in \tan y = \dfrac{{\sin \dfrac{\pi }{4} + \sin \dfrac{\theta }{2}}}{{\cos \dfrac{\pi }{4} + \cos \dfrac{\theta }{2}}} , we get \therefore$\tan y = \dfrac{{2\sin (\dfrac{\pi }{8} + \dfrac{\theta }{4})\cos (\dfrac{\pi }{8} - \dfrac{\theta }{4})}}{{2\cos (\dfrac{\pi }{8} + \dfrac{\theta }{4})\cos (\dfrac{\pi }{8} - \dfrac{\theta }{4})}}$
By simplifying we get , $\tan y = \dfrac{{\sin (\dfrac{\pi }{8} + \dfrac{\theta }{4})}}{{\cos (\dfrac{\pi }{8} + \dfrac{\theta }{4})}}$ hence we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$
$\Rightarrow$ $\tan y = \tan (\dfrac{\pi }{8} + \dfrac{\theta }{4})$
$\Rightarrow y = \dfrac{\pi }{8} + \dfrac{\theta }{4} \\ \Rightarrow 4y = \dfrac{\pi }{2} + \theta \\$
Hence according to question we have to find $\sin 4y$, so
$\sin 4y = \sin (\dfrac{\pi }{2} + \theta )$
Hence , $(\dfrac{\pi }{2} + \theta )$ lies in second quadrant , where $\sin$ is positive , hence it is equal to $\cos \theta$ and initially we have assumed $\cos \theta$$= x$.
$\therefore \sin 4y = x$