# Sec thita -1÷ sec thita +1 =(sin thita ÷ 1+ costhita)^2 ?

May 22, 2018

#### Explanation:

We need

$\sec \theta = \frac{1}{\cos} \theta$

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

Therefore the

$L H S = \frac{\sec \theta - 1}{\sec \theta + 1}$

$= \frac{\frac{1}{\cos} \theta - 1}{\frac{1}{\cos} \theta + 1}$

$= \frac{1 - \cos \theta}{1 + \cos \theta}$

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

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

${\sin}^{2} \frac{\theta}{1 + \cos \theta} ^ 2$

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

$= R H S$

$Q E D$

May 22, 2018

$L H S = \frac{\sec x - 1}{\sec x + 1}$

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

$= \frac{1 - \cos x}{1 + \cos x} \cdot \left[\frac{1 + \cos x}{1 + \cos x}\right]$

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

May 22, 2018

Explanation in below

#### Explanation:

$\frac{\sec x - 1}{\sec x + 1}$

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

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

=${\left(\tan x\right)}^{2} / {\left(\sec x + 1\right)}^{2}$

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

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

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

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