Question

For any acute angle $\theta ,$ find the value of the expression: $\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}$

Answer 1

Given: $\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}$ and $\theta$ is an acute angle.

Now, rationalising the denominator, we get:

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

$=\sqrt{\frac{(1+\sin \theta {)}^2}{(1-{\sin }^2\theta )}}$

Using the identity: ${\cos }^2\theta +{\sin }^2\theta =1$

$\sqrt{\frac{(1+\sin \theta {)}^2}{(1-{\sin }^2\theta )}}=\sqrt{\frac{(1+\sin \theta {)}^2}{{\cos }^2\theta }}$

Since, $\theta$ is an acute angle, both $\sin \theta \&\cos \theta$ will be positive.

$\Rightarrow \sqrt{\frac{(1+\sin \theta {)}^2}{{\cos }^2\theta }}=\frac{(1+\sin \theta )}{\cos \theta }$

$\Rightarrow \frac{(1+\sin \theta )}{\cos \theta }=\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=\sec \theta +\tan \theta$