# How to prove?

## $\frac{1 + \sec x}{{\tan}^{2} x} = \frac{\cos x}{1 - \cos x}$

Mar 3, 2018

$= L . H . S$

$= \frac{1 + \sec x}{{\tan}^{2} x}$

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

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

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

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

$= \frac{\cancel{\textcolor{b l u e}{\left(\cos x + 1\right)}} \cos x}{\cancel{\textcolor{b l u e}{\left(1 + \cos x\right)}} \left(1 - \cos x\right)}$

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

$= R . H . S \textcolor{g r e e n}{\left[P r o v e d .\right]}$

Mar 3, 2018

See below

#### Explanation:

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

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

$\frac{1 + \sec x}{\left(1 + \sec x\right) \left(1 - \sec x\right)} = \cos \frac{x}{1 - \cos x}$

$\frac{1}{1 - \sec x} = \cos \frac{x}{1 - \cos x}$

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

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

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