Question

under root seca-1/seca+1 + under root seca+1/seca-1 = 2coseca

can somebody answer this?

Answer 1

$$\begin{gathered} \frac{\sqrt{\sec A-1}}{\sqrt{\sec A+1}}+\frac{\sqrt{\sec A+1}}{\sqrt{\sec A-1}}=2CosecA \\ \mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{{\left(\sqrt{\sec \mathrm{A}-1}\right)}^2+{\left(\sqrt{\sec \mathrm{A}+1}\right)}^2}{\left(\sqrt{\sec \mathrm{A}+1}\right)\left(\sqrt{\sec \mathrm{A}-1}\right)} \\ =\frac{\sec \mathrm{A}-1+\sec \mathrm{A}+1}{\left(\sqrt{{\sec }^2\mathrm{A}-1}\right)} \\ =\frac{2\sec \mathrm{A}}{\left(\sqrt{{\tan }^2\mathrm{A}}\right)} \\ =\frac{2\mathrm{secA}}{\mathrm{tanA}} \\ =\frac{2}{\mathrm{cosA}}\times \frac{\mathrm{cosA}}{\mathrm{sinA}} \\ =\frac{2}{\mathrm{sinA}}=2\mathrm{cosecA}\left(\mathrm{answer}\right) \end{gathered}$$