Question

Show that $\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$

Answer 1

$\begin{array}{c}Wehave\\ L.H.S=\cos 2x={\cos }^{2}x-{\sin }^{2}x\\ =\frac{{\cos }^{2}x-{\sin }^{2}x}{1}\\ =\frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x+{\sin }^{2}x}\\ =\frac{{\cos }^{2}x\left\{1-\frac{{\sin }^{2}x}{{\cos }^{2}x}\right\}}{{\cos }^{2}x\left\{1+\frac{{\sin }^{2}x}{{\cos }^{2}x}\right\}}\\ =\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}=R.H.S\\ Hence,\\ L.H.S=R.H.S\end{array}$