Question

If \( ab < 1 \) and \( \cos^{-1}\left(\frac{1-a^2}{1+a^2}\right) + \cos^{-1\left(\frac{1-b^2}{1+b^2}\right) = 2 \tan^{-1} x \), then \( x \) is equal to:}

• A. \( \frac{a}{1+ab} \)
• B. \( \frac{a}{1-ab} \)
• C. \( \frac{a-b}{1+ab} \)
• D. \( \frac{a+b}{1+ab} \)
• E. \( \frac{a+b}{1-ab} \)

Hint:

Always remember the $2\tan^{-1}$ identities; they are the bridge between tan, sin, and cos inverse functions.

Answer 1

Answer: (E)

Concept: We utilize the standard inverse trigonometric identity: \[ 2 \tan^{-1} t = \cos^{-1}\left(\frac{1-t^2}{1+t^2}\right) \text{ for } t \ge 0 \] Additionally, the sum formula for tangent inverse is: \[ \tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \]

Step 1: Simplify the left-hand side using identities.
Substitute \( \cos^{-1}\left(\frac{1-a^2}{1+a^2}\right) = 2 \tan^{-1} a \) and \( \cos^{-1}\left(\frac{1-b^2}{1+b^2}\right) = 2 \tan^{-1} b \). The equation becomes: \[ 2 \tan^{-1} a + 2 \tan^{-1} b = 2 \tan^{-1} x \]

Step 2: Divide by 2 and apply the sum formula.
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} x \] \[ \tan^{-1}\left(\frac{a+b}{1-ab}\right) = \tan^{-1} x \]

Step 3: Extract x.
Comparing both sides: \[ x = \frac{a+b}{1-ab} \]