Question

Prove each of the following identities:

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

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{LHS}=\sqrt{\frac{1+\sin \theta }{1-\sin \theta }} \\ =\sqrt{\frac{\left(1+\sin \theta \right)}{\left(1-\sin \theta \right)}\times \frac{\left(1+\sin \theta \right)}{\left(1+\sin \theta \right)}} \\ =\sqrt{\frac{{\left(1+\sin \theta \right)}^2}{1-{\sin }^2\theta }} \\ =\sqrt{\frac{{\left(1+\sin \theta \right)}^2}{{\cos }^2\theta }} \\ =\frac{1+\sin \theta }{\cos \theta } \\ =\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta } \\ =\left(\sec \theta +\tan \theta \right) \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }} \\ =\sqrt{\frac{\left(1-\cos \theta \right)}{\left(1+\cos \theta \right)}\times \frac{\left(1-\cos \theta \right)}{\left(1-\cos \theta \right)}} \\ =\sqrt{\frac{{\left(1-\cos \theta \right)}^2}{1-{\cos }^2\theta }} \\ =\sqrt{\frac{{\left(1-\cos \theta \right)}^2}{{\sin }^2\theta }} \\ =\frac{1-\cos \theta }{\sin \theta } \\ =\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta } \\ =\left(\mathrm{cosec}\theta -\cot \theta \right) \\ =\mathrm{RHS} \end{gathered}$$

$\begin{array}{l}\left(\text{iii}\right)\text{LHS=}\sqrt{\frac{1+\text{cos}\theta }{1-\text{cos}\theta }}+\sqrt{\frac{1-\text{cos}\theta }{1+\text{cos}\theta }}\\ \text{=}\sqrt{\frac{{(1+\text{cos}\theta )}^2}{(1-\text{cos}\theta )(1+\text{cos}\theta )}}+\sqrt{\frac{{(1-\text{cos}\theta )}^2}{(1+\text{cos}\theta )(1-\text{cos}\theta )}}\\ \text{=}\sqrt{\frac{{(1+\text{cos}\theta )}^2}{(1-{\text{cos}}^2\theta )}}+\sqrt{\frac{{(1-\text{cos}\theta )}^2}{(1-{\text{cos}}^2\theta )}}\\ \text{=}\sqrt{\frac{{(1+\text{cos}\theta )}^2}{{\text{sin}}^2\theta }}+\sqrt{\frac{{(1-\text{cos}\theta )}^2}{{\text{sin}}^2\theta }}\\ \text{=}\frac{(1+\text{cos}\theta )}{\text{sin}\theta }+\frac{(1-\text{cos}\theta )}{\text{sin}\theta }\\ \text{=}\frac{1+\text{cos}\theta \text{+1}-\text{cos}\theta }{\text{sin}\theta }\\ \text{=}\frac{2}{\text{sin}\theta }\\ \text{=2cosec}\theta \\ \text{=RHS}\end{array}$