Question:
$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$ is equal to
(a) $\sec \theta+\tan \theta$
(b) $\sec \theta-\tan \theta$
(c) $\sec ^{2} \theta+\tan ^{2} \theta$
(d) $\sec ^{2} \theta-\tan ^{2} \theta$
Solution:
The given expression is $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$.
Multiplying both the numerator and denominator under the root by $(1+\sin \theta)$, we have
$\sqrt{\frac{(1+\sin \theta)(1+\sin \theta)}{(1+\sin \theta)(1-\sin \theta)}}$
$\sqrt{(1+\sin \theta)(1-\sin \theta)}$
$=\sqrt{\frac{(1+\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}$
$=\sqrt{\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}}$
$=\frac{1+\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}$
$=\sec \theta+\tan \theta$
Therefore, the correct option is (a)