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