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Question

Mathematics Question on Inverse Trigonometric Functions

If sin⁡−1x−cos⁡−1x=π6,\sin^{-1}x-\cos^{-1} x=\frac{\pi}{6}, then xx is

A

12\frac{1}{2}

B

32\frac{\sqrt3}{2}

C

−12-\frac{1}{2}

D

none of these.

Answer

32\frac{\sqrt3}{2}

Explanation

Solution

sin−1x−cos−1x=π6sin^{-1}x-cos^{-1} x = \frac{\pi}{6} ⇒π2−cos−1x−cos−1x=π6\Rightarrow \frac{\pi}{2} - cos^{-1}x-cos^{-1}x = \frac{\pi}{6} ⇒2 cos−1x=π2−π6 \Rightarrow2\,cos^{-1} x = \frac{\pi}{2} -\frac{\pi}{6} =2π6−π3= \frac{2\pi}{6} -\frac{\pi}{3} ⇒cos−1=π6\Rightarrow cos^{-1} = \frac{\pi}{6} ∴x=cosπ6\therefore x= cos \frac{\pi}{6} =32 = \frac{\sqrt{3}}{2}