Question

If $x=\sin \left(2{\tan }^{-1}\right(1/3\left)\right)+\cos \left({\tan }^{-1}\right(2\sqrt{2}\left)\right)$ then $\frac{1}{(1-x{)}^3}+\frac{1}{(1-x{)}^2}+\frac{1}{(1-x)}$ is equal to

Answer 1

$x=\sin \left(2{\tan }^{-1}\right(\frac{1}{3}\left)\right)+\cos \left({\tan }^{-1}2\sqrt{2}\right)$
$=2\sin \left({\tan }^{-1}\right(\frac{1}{3}\left)\cos \right({\tan }^{-1}\left(\frac{1}{3}\right)+\cos \left({\cos }^{-1}\frac{1}{3}\right)$
$=2\sin \left({\sin }^{-1}\frac{1}{\sqrt{10}}\right)\cos \left({\cos }^{-1}\frac{3}{\sqrt{10}}\right)+\frac{1}{3}$
$=\frac{2.3}{\sqrt{10}.\sqrt{10}}+\frac{1}{3}$
$=\frac{6}{10}+\frac{1}{3}$
$=\frac{14}{15}$
Hence $1-x$
$=\frac{1}{15}$
Hence $(15{)}^3+{15}^2+15$
$=15({15}^2+15+1)$
$=15\left(241\right)$
$=3615$