Question

Whole root of secA-1 / secA+1 +whole root of secA+1 / secA-1 = 2cosecA

Answer 1

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