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Prove the following trigonometric identities:$ \frac{(1+\sin \theta)^{2}+(1-\sin \theta)^{2}}{2 \cos ^{2} \theta}=\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta} $
To do:
We have to prove that \( \frac{(1+\sin \theta)^{2}+(1-\sin \theta)^{2}}{2 \cos ^{2} \theta}=\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta} \).
Solution:
We know that,
$\cos ^{2} \theta+\sin^2 \theta=1$.......(i)
Therefore,
$\frac{(1+\sin \theta)^{2}+(1-\sin \theta)^{2}}{2 \cos ^{2} \theta}=\frac{1+\sin^2 \theta+2\sin \theta+1-2\sin \theta+\sin^2 \theta}{2\cos^2 \theta}$
$=\frac{2+2\sin^2 \theta}{2\cos^2 \theta}$
$=\frac{2(1+\sin^2 \theta)}{2\cos^2 \theta}$
$=\frac{1+\sin^2 \theta}{\cos^2 \theta}$
$=\frac{1+\sin^2 \theta}{1-\sin^2 \theta}$ [From (i)]
Hence proved.
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