What is \tan^{-1}(\frac{a}{b})-\tan^{-1}(\frac{a-b}{a+b}) equal to ?
- A. -\frac{\pi}{4}
- B. \frac{\pi}{4} ✓
- C. \tan^{-1}(\frac{a^{2}-b^{2}}{a^{2}+b^{2}})
- D. \tan^{-1}(\frac{2ab}{a^{2}+b^{2}})
Correct Answer: B. \frac{\pi}{4}
Explanation
Rewrite the second term: \tan^{-1}(\frac{a-b}{a+b}) = \tan^{-1}\left(\frac{\frac{a}{b}-1}{1+\frac{a}{b}}\right). Using the identity \tan^{-1}(\frac{x-y}{1+xy}) = \tan^{-1}x - \tan^{-1}y, this equals \tan^{-1}(\frac{a}{b}) - \tan^{-1}(1). Subtracting this from the first term leaves exactly \tan^{-1}(1) = \frac{\pi}{4}.