Question

If \(\sin^{-1} x + \cot^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2}\), then value of \(x\) will be

• A. 0
• B. \(\frac{1}{\sqrt{5}}\)
• C. \(\frac{2}{\sqrt{5}}\)
• D. \(\frac{\sqrt{3}}{2}\)

Hint:

Use identity: \(\cot^{-1}x = \tan^{-1}(1/x)\) and \(\frac{\pi}{2}-\tan^{-1}x = \tan^{-1}(1/x)\).

Answer 1

The Correct Option is B

Step 1: Evaluate \(\cot^{-1}(1/2)\).
\[ \cot^{-1}\left(\frac{1}{2}\right) = \tan^{-1}(2) \]

Step 2: Substitute.
\[ \sin^{-1}x + \tan^{-1}(2) = \frac{\pi}{2} \] \[ \Rightarrow \sin^{-1}x = \frac{\pi}{2} - \tan^{-1}(2) \] \[ \Rightarrow \sin^{-1}x = \tan^{-1}\left(\frac{1}{2}\right) \]

Step 3: Convert to sine.
Let \(\theta = \tan^{-1}(1/2)\) \[ \sin \theta = \frac{1}{\sqrt{1^2+2^2}} = \frac{1}{\sqrt{5}} \] \[ \Rightarrow x = \frac{1}{\sqrt{5}} \]