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(2\, tan^{-1}\,2\right) $
$ = sin\, 2\theta$ where $tan \theta = 2 $
$ = \frac{2\,tan\,\theta}{1+tan^{2} \,\theta} = \frac{2\left(2\right)}{1+4} = \frac{4}{5} $
$ y= sin\left(\frac{1}{2} tan^{-1} \frac{4}{3}\right) $
$ = sin \frac{\phi}{2}$
where $tan \phi = \frac{4}{3} $
$ \therefore cos \phi = \frac{3}{5} $
$ \therefore 1-2 sin^{2} \frac{\phi}{2} = \frac{3}{5}$
$ \Rightarrow 2sin^{2} \frac{\phi}{2} = \frac{2}{5} $
$ \Rightarrow sin^{2} \frac{\phi}{2} = \frac{1}{5}$
$\Rightarrow sin \frac{\phi}{2} = \pm \frac{1}{\sqrt{5}} $
$\therefore y= \pm\frac{1}{\sqrt{5}} $
$ \therefore x > y $ and $y^{2}=1-x $