Question
Prove the following identities:
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=(\sec\theta+\tan\theta)$
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=(\sec\theta+\tan\theta)$
✓
Answer
$\text{LHS}=\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}$
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}\times\frac{1+\sin\theta}{1+\sin\theta}}$
$\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}$
$\sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}$
$=\frac{1+\sin\theta}{\cos\theta}$
$=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}$
$=\sec\theta+\tan\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}\times\frac{1+\sin\theta}{1+\sin\theta}}$
$\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}$
$\sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}$
$=\frac{1+\sin\theta}{\cos\theta}$
$=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}$
$=\sec\theta+\tan\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
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