Question

If $x=\sin \left(2{\tan }^{-1}2\right),y=\sin \left(\frac{1}{2}{\tan }^{-1}\frac{4}{3}\right)$, then

• A. $x=1-y$
• B. $x^2=1-y$
• C. $x^2=1+y$
• D. $y^2=1-x$

Answer 1

Answer: (D)

$y^2=1-x$

Given, $x=\sin \left(2{\tan }^{-1}2\right),y=\sin \left(\frac{1}{2}{\tan }^{-1}\frac{4}{3}\right)$

$x=\sin 2\theta$ where $\tan \theta =2$

$\Rightarrow x=\frac{2\tan \theta }{1+{\tan }^2\theta }=\frac{4}{1+4}=\frac{4}{5}$

If $\alpha ={\tan }^{-1}\frac{4}{5},y=\sin \left(\frac{\alpha }{2}\right)=\frac{1}{\sqrt{2}}\sqrt{1-\cos \alpha }$

$=\frac{1}{\sqrt{2}}\sqrt{1-\frac{3}{5}}=\frac{1}{\sqrt{5}}$

$\therefore y^2=\frac{1}{5}\Rightarrow y^2=1-x$