Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given $x=\sin \left(2{\tan }^{-1}\alpha \right)=\sin \left({\tan }^{-1}\frac{2\alpha }{1-{\alpha }^2}\right)=\sin \left({\sin }^{-1}\frac{2\alpha }{1+{\alpha }^2}\right)=\frac{2\alpha }{1+{\alpha }^2} \cdots \left(\mathrm{i}\right)$

and $y=\sin \left(\frac{1}{2}{\tan }^{-1}\frac{4}{3}\right)=\sin \left({\tan }^{-1}\frac{1}{2}\right)=\sin \left({\sin }^{-1}\frac{1}{\sqrt{5}}\right)=\frac{1}{\sqrt{5}} ...\left(\mathrm{ii}\right)$

Now, $y^2=1-x$

$\Rightarrow \frac{1}{5}=1-\frac{2\alpha }{1+{\alpha }^2}$ (from $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$)

$\Rightarrow 1+{\alpha }^2=5+5{\alpha }^2-10\alpha$

$\Rightarrow 2{\alpha }^2-5\alpha +2=0$

$\therefore \alpha =2,\frac{1}{2}$

$\therefore \sum _{\alpha \in S}16{\alpha }^3=16\times 2^3+16\times \frac{1}{2^3}=16\left(\frac{65}{8}\right)$

$=130$