Question

If $\frac{\mathrm{\pi }}{2}<\mathrm{\theta }<\frac{3\mathrm{\pi }}{2},\mathrm{then}\sqrt{\frac{1-\mathrm{sin\theta }}{1+\mathrm{sin\theta }}}$ is equal to
(a) sec θ − tan θ
(b) sec θ + tan θ
(c) tan θ − sec θ
(d) none of these

Answer 1

(c) tan θ − sec θ

$$\begin{gathered} \sqrt{\frac{1-\mathrm{sin\theta }}{1+\mathrm{sin\theta }}} \\ =\sqrt{\frac{\left(1-\mathrm{sin\theta }\right)\left(1-\mathrm{sin\theta }\right)}{\left(1+\mathrm{sin\theta }\right)\left(1-\mathrm{sin\theta }\right)}} \\ =\sqrt{\frac{{\left(1-\mathrm{sin\theta }\right)}^2}{1-{\sin }^2\theta }} \\ =\sqrt{\frac{{\left(1-\mathrm{sin\theta }\right)}^2}{{\cos }^2\theta }} \\ =\frac{\left(1-\mathrm{sin\theta }\right)}{-\mathrm{co}s\theta }\left[\mathrm{as},\frac{\mathrm{\pi }}{2}<\theta <\frac{3\mathrm{\pi }}{2},\mathrm{so}\cos \theta \mathrm{will}\mathrm{be}\mathrm{negative}\right] \\ =-\left(sec\theta -\tan \theta \right) \\ =-sec\theta +\tan \theta \end{gathered}$$