Prove the following trigonometric identities:$ \frac{1+\cos A}{\sin ^{2} A}=\frac{1}{1-\cos A} $

To do:

We have to prove that \( \frac{1+\cos A}{\sin ^{2} A}=\frac{1}{1-\cos A} \).

Solution:

We know that,

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

$(a+b)(a-b)=a^2-b^2$.........(ii)

Therefore,

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

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

$=\frac{1+\cos A}{(1+\cos A)(1-\cos A)}$        [From (ii)]          

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

Hence proved.