1+sin θ1−sin θ−−−−−√ is equal to
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)