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)

$x>y$ and $y^2=1-x$

Answer 2

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 $y^{2}=1-x$