Question

Prove that $\cos 2x=\frac{(1-{\tan }^{2}x)}{(1+{\tan }^{2}x)}$

Answer 1

$\cos 2x={\cos }^{2}x–{\sin }^{2}x$

$=\frac{({\cos }^{2}x–{\sin }^{2}x)}{1}$

We know that, ${\cos }^2\theta +{\sin }^2\theta =1$

$=\frac{({\cos }^{2}x–{\sin }^{2}x)}{\left({\cos }^{2}x+{\sin }^{2}x\right)}$

$=\frac{{\cos }^{2}x\left(1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}\right)}{{\cos }^{2}x\left(1+{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}\right)}$

$=\frac{\left(1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}\right)}{\left(1+{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}\right)}$

$=\frac{\left(1-{\displaystyle {\tan }^{2}x}\right)}{\left(1+{\displaystyle {\tan }^{2}x}\right)}$.

Hence, it is proved that $\cos 2x=\frac{(1-{\tan }^{2}x)}{(1+{\tan }^{2}x)}$.