Question

If $\sin^{-1}\left(\dfrac{1}{2}\right) = \tan^{-1}x$, find the value of $x$.

Hint:

Use standard values of trigonometric functions: $\sin^{-1}\left(\tfrac{1}{2}\right) = \tfrac{\pi}{6}$, $\cos^{-1}\left(\tfrac{1}{2}\right) = \tfrac{\pi}{3}$, etc.

Answer 1

Step 1: Simplify the inverse sine.
\[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] So, \[ \tan^{-1}x = \frac{\pi}{6} \]

Step 2: Apply tangent function.
Taking tangent on both sides: \[ x = \tan\left(\frac{\pi}{6}\right) \] \[ x = \frac{1}{\sqrt{3}} \]

Final Answer: \[ \boxed{\dfrac{1}{\sqrt{3}}} \]