Question

\[ \sin^{-1} x - \cos^{-1} 2x = \sin^{-1} \left(\frac{\sqrt{3}}{2}\right) - \cos^{-1} \left(\frac{\sqrt{3}}{2}\right) \]

Then, \[ \tan^{-1} x + \tan^{-1} \left(\frac{x}{x+1}\right) = ? \]

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{\pi}{2}\)]

Hint:

The tangent addition formula, \( \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right) \), is very useful for summing angles in inverse tangent form.

Answer 1

The Correct Option is B

First, recognize that \( \sin^{-1} x \) and \( \cos^{-1} 2x \) are trigonometric inverses, and equating their differences to specific values can be simplified if we match these values to known angles. Given that \( \sin^{-1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \) and \( \cos^{-1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \), we find: \[ \sin^{-1} x - \cos^{-1} 2x = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \] Now, solve for \(x\) knowing that: \[ \sin^{-1} x = \frac{\pi}{6} + \cos^{-1} 2x \] And since \( \sin^{-1} x = \frac{\pi}{6} \) implies \( x = \frac{1}{2} \). Using this in the expression \( \tan^{-1} x + \tan^{-1} \left(\frac{x}{x+1}\right) \): \[ \tan^{-1} \frac{1}{2} + \tan^{-1} \left(\frac{1/2}{3/2}\right) = \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} \] Using the tangent addition formula: \[ \tan^{-1} \frac{1/2} + \tan^{-1} \frac{1}{3} = \tan^{-1} \left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} \right) = \tan^{-1} 1 = \frac{\pi}{4} \]