Question

If sin⁻¹ x − cos⁻¹ x = $\frac{\mathrm{\pi }}{6}$, then x =
(a) $\frac{1}{2}$

(b) $\frac{\sqrt{3}}{2}$

(c) $-\frac{1}{2}$

(d) none of these

Answer 1

(b) $\frac{\sqrt{3}}{2}$
We know that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}$.

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