Prove the following trigonometric identities:$ \frac{1+\cos A}{\sin A}=\frac{\sin A}{1-\cos A} $

To do:

We have to prove that \( \frac{1+\cos A}{\sin A}=\frac{\sin A}{1-\cos A} \).

Solution:

We know that,

$\sin ^{2} A+\cos^2 A=1$.......(i)

Therefore,

$\frac{1+\cos A}{\sin A}=\frac{1+\cos A}{\sin A} \times \frac{1-\cos A}{1-\cos A}$     (Multiplying and dividing by $1-\cos A$)

$=\frac{1+\cos A(1-\cos A)}{\sin A(1-\cos A)}$

$=\frac{1^2-\cos^2 A}{\sin A(1-\cos A)}$
$=\frac{\sin^2 A}{\sin A(1-\cos A)}$              [From (i)]

$=\frac{\sin A}{1-\cos A}$

Hence proved.