Question
Prove that:
$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta$
$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta$
✓
Answer
Taking LHS
$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
$=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \times \frac{\sqrt{1-\sin \theta}}{\sqrt{1-\sin \theta}}$
$=\frac{1-\sin \theta}{\sqrt{(1+\sin \theta)(1-\sin \theta)}}$
$=\frac{1-\sin \theta}{\sqrt{1-\sin ^2 \theta}}\left[(a+b)(a-b)= a ^2- b ^2\right]$
$=\frac{1-\sin ^\theta}{\sqrt{\cos ^2 \theta}}\left[ As , \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\frac{1-\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$
$=\sec \theta-\tan \theta$
= RHS
Proved !
$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
$=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \times \frac{\sqrt{1-\sin \theta}}{\sqrt{1-\sin \theta}}$
$=\frac{1-\sin \theta}{\sqrt{(1+\sin \theta)(1-\sin \theta)}}$
$=\frac{1-\sin \theta}{\sqrt{1-\sin ^2 \theta}}\left[(a+b)(a-b)= a ^2- b ^2\right]$
$=\frac{1-\sin ^\theta}{\sqrt{\cos ^2 \theta}}\left[ As , \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\frac{1-\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$
$=\sec \theta-\tan \theta$
= RHS
Proved !
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