If sec⁻¹ x = cosec⁻¹ y, show that 1/(x²) + 1/(y²) = 1

If $\sec^{-1} x = \operatorname{cosec}^{-1} y$, show that $\frac{1}{x^2} + \frac{1}{y^2} = 1$

Given: $\sec^{-1} x = \operatorname{cosec}^{-1} y$

To show: $\frac{1}{x^2} + \frac{1}{y^2} = 1$

Answer

Put both inverse functions equal to $\theta$; then $x=\sec\theta$ and $y=\operatorname{cosec}\theta$, so $\frac{1}{x^2}+\frac{1}{y^2} = \cos^2\theta+\sin^2\theta = 1$.

1. Let $\sec^{-1}x = \operatorname{cosec}^{-1}y = \theta$ (given).
2. Then $x = \sec\theta$ and $y = \operatorname{cosec}\theta$ (definition of inverse functions).
3. So $\frac{1}{x^2} = \cos^2\theta$ and $\frac{1}{y^2} = \sin^2\theta$ (reciprocal identities).
4. $\frac{1}{x^2}+\frac{1}{y^2} = \cos^2\theta + \sin^2\theta$ (adding).
5. $\cos^2\theta+\sin^2\theta = 1$ (Pythagorean identity).
6. Hence $\frac{1}{x^2}+\frac{1}{y^2} = 1$.