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

To find the value of \(x\) in the equation \(\sin^{-1} x + \cot^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2}\), let's break it down step by step:

1. First, understand that the expression \(\sin^{-1} x + \theta = \frac{\pi}{2}\) implies \(\sin^{-1} x\) and \(\theta\) are complementary angles. Therefore, \(\theta = \frac{\pi}{2} - \sin^{-1} x\).
2. In the given equation, \(\theta\) is \(\cot^{-1}\left(\frac{1}{2}\right)\). So, we have: \[ \cot^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2} - \sin^{-1} x \]
3. Using the property \(\cot^{-1} y = \tan^{-1} \left(\frac{1}{y}\right)\), we can rewrite: \[ \cot^{-1}\left(\frac{1}{2}\right) = \tan^{-1} (2) \]
4. Substitute back to the equation: \[ \tan^{-1} (2) = \frac{\pi}{2} - \sin^{-1} x \]
5. Now rearrange the equation: \[ \sin^{-1} x = \frac{\pi}{2} - \tan^{-1} (2) \]
6. Recall the identity \(\tan^{-1} (a) + \sin^{-1} (b) = \frac{\pi}{2}\): \[ \sin^{-1} \left(\frac{1}{\sqrt{1+a^2}}\right) = \sin^{-1} x \] where \(a = 2\).
7. Therefore, we substitute \(a = 2\) into the identity: \[ x = \frac{1}{\sqrt{1+2^2}} = \frac{1}{\sqrt{5}} \]

Thus, the value of \(x\) is \(\frac{1}{\sqrt{5}}\). Therefore, the correct answer is: \(\frac{1}{\sqrt{5}}\).