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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.
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