Question

Prove the following trigonometric identities.

${\displaystyle \frac{\sin \theta }{1-\cos \theta }= \cos ec\theta +\cot \theta }$

Answer 1

We have to prove  ${\displaystyle \frac{\sin \theta }{1-\cos \theta }=\cos ec\theta +\cot \theta }$

We know that ${\displaystyle {\sin }^2\theta ={\cos }^2\theta =1}$

${\displaystyle \frac{\sin \theta }{1-\cos \theta }=\frac{\sin \theta (1+\cos \theta )}{1-{\cos }^2\theta }}$

${\displaystyle =\frac{\sin \theta (1+ \cos \theta )}{1- {\cos }^2\theta }}$`

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

${\displaystyle =\frac{1+\cos \theta }{\sin \theta }}$

${\displaystyle =\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }}$

${\displaystyle =\cos ec\theta +\cot \theta }$