Question

$\sqrt{\left(\frac{1+\sin \theta }{1-\sin \theta }\right)}$ =

• A. $\tan \theta -sec\theta$
• B. $\tan \theta +sec\theta$
• C. $sec\theta -\tan \theta$
• D. none of these

Answer 1

The correct option is B

$\tan \theta +sec\theta$

Step 1:

Let $\sqrt{\left(\frac{1+\sin \theta }{1-\sin \theta }\right)}=x$

Step 2:

Multiply and divide the above equation by $(1+\sin \theta )$.

Step 3:

$$\begin{gathered} x=\sqrt{\left(\frac{1+\sin \theta }{1-\sin \theta }\right)\times \frac{1+\sin \theta }{1+\sin \theta }} \\ x=\sqrt{\left(\frac{{\left(1+\sin \theta \right)}^2}{1-{\sin }^2\theta }\right)} \\ U\sin g, \\ (1+\sin \theta )(1-\sin \theta )=1-{\sin }^2\theta ={\cos }^2\theta \\ x=\sqrt{\left(\frac{{\left(1+\sin \theta \right)}^2}{{\cos }^2\theta }\right)} \\ x=\frac{1+\sin \theta }{\cos \theta } \\ x=\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta } \\ x=sec\theta +\tan \theta \end{gathered}$$

Hence, option (B) is correct.