Question

If 2x = sec A and $\frac{2}{x}$ = tan A, prove that $\left(x^2-\frac{1}{x^2}\right)=\frac{1}{4}.$

Answer 1

$$\begin{gathered} \text{Given: 2}\mathrm{x}=\mathrm{secA} \\ \text{=>}\mathrm{x}=\frac{\mathrm{secA}}{2}...\left(\mathrm{i}\right) \\ \mathrm{and}\frac{2}{\mathrm{x}}=\mathrm{tanA} \\ \text{=>}\frac{1}{\mathrm{x}}=\frac{\mathrm{tanA}}{2}...\left(\mathrm{ii}\right) \\ \therefore \mathrm{x}+\frac{1}{\mathrm{x}}=\frac{\mathrm{secA}}{2}+\frac{\mathrm{tanA}}{2}[\because \mathrm{From}(\mathrm{i})\mathrm{and}(\mathrm{ii}\left)\right] \\ \text{Also,}\mathrm{x}-\frac{1}{\mathrm{x}}=\frac{\mathrm{secA}}{2}-\frac{\mathrm{tanA}}{2} \\ \therefore (\mathrm{x}+\frac{1}{\mathrm{x}})(\mathrm{x}-\frac{1}{\mathrm{x}})=(\frac{\mathrm{secA}}{2}+\frac{\mathrm{tanA}}{2})(\frac{\mathrm{secA}}{2}-\frac{\mathrm{tanA}}{2}) \\ \text{=>}{\mathrm{x}}^2-\frac{1}{{\mathrm{x}}^2}=\frac{1}{4}({\sec }^2\mathrm{A}-{\tan }^2\mathrm{A}) \\ \therefore {\mathrm{x}}^2-\frac{1}{{\mathrm{x}}^2}=\frac{1}{4}\times 1(\because {\sec }^2\mathrm{A}-{\tan }^2\mathrm{A}=1) \\ \text{=}\frac{1}{4} \\ \text{Hence proved.} \end{gathered}$$