Question

If $\frac{\mathrm{\pi }}{2}<\mathrm{\theta }<\mathrm{\pi },\mathrm{then}\sqrt{\frac{1-\mathrm{sin\theta }}{1+\mathrm{sin\theta }}}+\sqrt{\frac{1+\mathrm{sin\theta }}{1-\mathrm{sin\theta }}}$ is equal to
(a) 2 sec θ
(b) −2 sec θ
(c) sec θ
(d) −sec θ

Answer 1

(b) −2 sec θ

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