Question

How do you simplify $\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$?

Answer 1

Simplify the given expression:

The given expression is

$\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}}{1+\frac{{\sin }^{2}x}{{\cos }^{2}x}}$ $...\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{x}\mathbf{=}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{x}}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{x}}\right]$

$=\frac{{\displaystyle \frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x}}}{\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}}$

$=\frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x+{\sin }^{2}x}$ $...\left[\because cos2x=co{s}^{2}x-si{n}^{2}x;si{n}^{2}x+co{s}^{2}x=1\right]$

$=\frac{\cos 2x}{1}$

$=\cos 2x$

Hence, $\frac{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{x}}{\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{x}}$ can be simplified into $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{x}$.