Question

If $\frac{\pi }{2}<\theta <\frac{3\pi }{2}$, then $\sqrt{\frac{1-sin\theta }{1+sin\theta }}$ is equal to

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

Answer 1

The correct option is C

$tan\theta -sec\theta$

$tan\theta -sec\theta$

$\sqrt{\frac{1-sin\theta }{1+sin\theta }}$

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

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

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

$=\frac{(1-sin\theta )}{-cos\theta }$ [as, $\frac{\pi }{2}<\theta <\frac{3\pi }{2}$, so $cos\theta$ will be negative]

$=-(sec\theta -tan\theta )$

$=-sec\theta +tan\theta$