Question
Prove the trigonometric identity: $\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}=2 \sec ^2 \theta$
✓
Answer
$\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}=2 \sec ^2 \theta$
$\text { L.H.S. }=\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}$
$=\frac{1+\sin \theta+1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)}=\frac{2}{1-\sin ^2 \theta}$
$=\frac{2}{\cos ^2 \theta}\left[\because 1-\sin ^2 \theta=\cos ^2 \theta\right]$
$=2 \sec ^2 \theta\left[\because \sec (x)=\frac{1}{\cos (x)}\right]$
$=\text { R.H.S. Proved }$
$\text { L.H.S. }=\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}$
$=\frac{1+\sin \theta+1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)}=\frac{2}{1-\sin ^2 \theta}$
$=\frac{2}{\cos ^2 \theta}\left[\because 1-\sin ^2 \theta=\cos ^2 \theta\right]$
$=2 \sec ^2 \theta\left[\because \sec (x)=\frac{1}{\cos (x)}\right]$
$=\text { R.H.S. Proved }$
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