Question

If sin^-1x - cos^-1x = $\frac{\mathrm{\pi }}{6}$, then x = _________________________.

Answer 1

Given: ${\sin }^{-1}x-{\cos }^{-1}x=\frac{\mathrm{\pi }}{6}$ .....(1)

We know

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}$ .....(2)

Adding (1) and (2), we get

$$\begin{gathered} 2{\sin }^{-1}x=\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{2} \\ \Rightarrow 2{\sin }^{-1}x=\frac{4\mathrm{\pi }}{6} \\ \Rightarrow {\sin }^{-1}x=\frac{\mathrm{\pi }}{3} \end{gathered}$$
$\Rightarrow x=\sin \frac{\mathrm{\pi }}{3}=\frac{\sqrt{3}}{2}$

If sin⁻¹x − cos⁻¹x = $\frac{\mathrm{\pi }}{6}$, then x = $\underline{\frac{\sqrt{3}}{2}}$.