Question

Question 21

$$\begin{gathered} 21.\left(\mathrm{i}\right)\sqrt{\frac{1+\sin \mathrm{\theta }}{1-\sin \mathrm{\theta }}}=\left(\sec \mathrm{\theta }+\tan \mathrm{\theta }\right) \\ \left(\mathrm{ii}\right)\sqrt{\frac{1-\cos \mathrm{\theta }}{1+\cos \mathrm{\theta }}}=\left(\mathrm{cosec}\mathrm{\theta }-\tan \mathrm{\theta }\right) \\ \left(\mathrm{iii}\right)\sqrt{\frac{1+\cos \mathrm{\theta }}{1-\cos \mathrm{\theta }}}+\sqrt{\frac{1-\cos \mathrm{\theta }}{1+\cos \mathrm{\theta }}}=2\mathrm{cosec}\mathrm{\theta } \end{gathered}$$

Answer 1

Dear student

$$\begin{gathered} 1)\sqrt{\frac{1+\mathrm{sin\theta }}{1-\mathrm{sin\theta }}} \\ =\sqrt{\frac{1+\mathrm{sin\theta }}{1-\mathrm{sin\theta }}}\times \sqrt{\frac{1+\mathrm{sin\theta }}{1+\mathrm{sin\theta }}} \\ =\frac{\sqrt{{\left(1+\mathrm{sin\theta }\right)}^2}}{\sqrt{1-{\sin }^2\mathrm{\theta }}} \\ =\frac{1+\mathrm{sin\theta }}{\sqrt{{\cos }^2\mathrm{\theta }}}\left[\mathrm{As}{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}=1\right] \\ =\frac{1+\mathrm{sin\theta }}{\mathrm{cos\theta }} \\ =\frac{1}{\mathrm{cos\theta }}+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }} \\ =\mathrm{sec\theta }+\mathrm{tan\theta } \end{gathered}$$
Similarly try other parts.
Regards