# Prove that cosB/(1-sinB)=secB+tanB?

Sep 10, 2016

#### Explanation:

$\cos \frac{B}{1 - \sin B}$

= $\cos \frac{B}{1 - \sin B} \times \frac{1 + \sin B}{1 + \sin B}$

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

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

= $\frac{1 + \sin B}{\cos} B$

= $\frac{1}{\cos} B + \sin \frac{B}{\cos} B$

= $\sec B + \tan B$