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)