Question

If $ \cos^{-1}(x) - \sin^{-1}(x) = \frac{\pi}{6} $, then find } $ x $.

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

Hint:

For inverse trigonometric functions, knowing that \( \cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2} \) is helpful for simplifying such equations.

Answer 1

The Correct Option is A

We are given: \[ \cos^{-1}(x) - \sin^{-1}(x) = \frac{\pi}{6} \] We know that: \[ \cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2} \] Now, substitute this into the given equation: \[ \frac{\pi}{2} - \sin^{-1}(x) - \sin^{-1}(x) = \frac{\pi}{6} \] Simplifying this: \[ 2 \sin^{-1}(x) = \frac{\pi}{3} \] Now, divide both sides by 2: \[ \sin^{-1}(x) = \frac{\pi}{6} \] Now, taking the sine of both sides: \[ x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \] 
Thus, the value of \( x \) is \( \frac{1}{2} \).