Question

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

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

Answer 1

Answer: (A)

Let $x=sin\left(2ta{n}^{-1}2\right)$
$=sin2\theta$ where $tan\theta =2$
$=\frac{2tan\theta }{1+ta{n}^2\theta }=\frac{2\left(2\right)}{1+4}=\frac{4}{5}$
$y=sin\left(\frac{1}{2}ta{n}^{-1}\frac{4}{3}\right)$
$=sin\frac{\varphi }{2}$
where $tan\varphi =\frac{4}{3}$
$\therefore cos\varphi =\frac{3}{5}$
$\therefore 1-2si{n}^2\frac{\varphi }{2}=\frac{3}{5}$
$\Rightarrow 2si{n}^2\frac{\varphi }{2}=\frac{2}{5}$
$\Rightarrow si{n}^2\frac{\varphi }{2}=\frac{1}{5}$
$\Rightarrow sin\frac{\varphi }{2}=\pm \frac{1}{\sqrt{5}}$
$\therefore y=\pm \frac{1}{\sqrt{5}}$
$\therefore x>y$ and $y^2=1-x$