# Question #98f3d

Nov 30, 2017

Eeeezy peeeezy....

#### Explanation:

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

$\frac{1 - {\sin}^{2} \theta}{1 - \sin \theta} - \frac{1 - \sin \theta}{1 - \sin \theta}$

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

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

$= \sin \theta$

GOOD LUCK

Nov 30, 2017

See below.

#### Explanation:

Not sure if you are asking to prove this, or you are looking to solve it as an equation for specific values of theta.

This is not an identity, so it can't be proved.

LHS

Let $\theta = \frac{\pi}{2}$

$\frac{{\cos}^{2} \left(\frac{\pi}{2}\right)}{1 - \sin \left(\frac{\pi}{2}\right)} - 1 = \frac{0}{1 - 1} - 1 =$( division by zero)

RHS:

$\sin \left(\frac{\pi}{2}\right) = 1$

$L H S \ne R H S$