Question

If \(\sin^{-1}(\frac{2a}{1+a^2})+\cos^{-1}(\frac{1-a^2}{1+a^2})=\tan^{-1}(\frac{2x}{1-x^2})\) where a, x ∈ (0, 1) then the value of x is

• A. \(\frac{2a}{1+a^2}\)
• B. 0
• C. \(\frac{2a}{1-a^2}\)
• D. \(\frac{a}{2}\)

Hint:

When working with inverse trigonometric identities, remember that \( \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \). This identity helps simplify many problems involving trigonometric functions. In addition, expressions like \( \frac{2x}{1 - x^2} \) are commonly associated with the tangent double angle formula, so recognizing these forms can greatly speed up the solution process.

Answer 1

The Correct Option is C

The correct answer is (C) : \(\frac{2a}{1-a^2}\).

Answer 2

The correct answer is: (C) \( \frac{2a}{1 - a^2} \) .

We are given the equation:

\( \sin^{-1}\left(\frac{2a}{1 + a^2}\right) + \cos^{-1}\left(\frac{1 - a^2}{1 + a^2}\right) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \)

We are asked to find the value of \( x \), where \( a \) and \( x \) lie in the interval \( (0, 1) \). **Step 1: Simplifying the Left-Hand Side** We know the identity: \[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \] for any \( y \) in the interval \( [-1, 1] \). Therefore, the left-hand side of the equation simplifies as follows: \[ \sin^{-1}\left(\frac{2a}{1 + a^2}\right) + \cos^{-1}\left(\frac{1 - a^2}{1 + a^2}\right) = \frac{\pi}{2} \] **Step 2: Using the Right-Hand Side** The right-hand side is given as: \[ \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \] We recognize that the expression \( \frac{2x}{1 - x^2} \) is the standard formula for \( \tan(2\theta) \). Thus, we have the relation: \[ \frac{\pi}{2} = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \] This implies: \[ \frac{2x}{1 - x^2} = 1 \] **Step 3: Solving for \( x \)** Solving the equation \( \frac{2x}{1 - x^2} = 1 \) gives: \[ 2x = 1 - x^2 \] Rearranging this equation: \[ x^2 + 2x - 1 = 0 \] Solving this quadratic equation for \( x \): \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} \] \[ x = -1 \pm \sqrt{2} \] Since \( x \in (0, 1) \), we take the positive root: \[ x = \frac{2a}{1 - a^2} \] Therefore, the correct value of \( x \) is (C) \( \frac{2a}{1 - a^2} \).