Question

How do you prove $se{c}^{2}x-{\tan }^{2}x=1$?

Answer 1

Proof of given relation:

$se{c}^{2}x-{\tan }^{2}x=1$

Here we have $LHS=se{c}^{2}x-{\tan }^{2}x$

Therefore,

$se{c}^{2}x-{\tan }^{2}x=\frac{1}{{\cos }^{2}x}-\frac{{\sin }^{2}x}{{\cos }^{2}x}$ $\left[\because se{c}^{2}x=\frac{1}{co{s}^{2}x};ta{n}^{2}x=\frac{si{n}^{2}x}{co{s}^{2}x}\right]$

$=\frac{1-{\sin }^{2}x}{{\cos }^{2}x}$

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

$=1$

$=RHS$

Hence, $\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathbf{-}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathbf{1}$ is proved.