Q.

$\begin{array}{l}\tan \mathrm{x}\left(1+{\tan }^2\mathrm{x}\right)=1-{\tan }^2\mathrm{x}\\ \tan \mathrm{x}=\frac{1-{\tan }^2\mathrm{x}}{1+{\tan }^2\mathrm{x}}=\cos 2\mathrm{x}\\ {\tan }^2\mathrm{x}={\cos }^{2}2\mathrm{x}\\ {\sin }^2\mathrm{x}={\cos }^2\mathrm{x}{\left(2{\cos }^2\mathrm{x}-1\right)}^2\\ 2{\cos }^6\mathrm{x}-2{\cos }^4\mathrm{x}+{\cos }^2\mathrm{x}=1/2\end{array}$