allery$\tan^{-1}\left(\tfrac{1}{2}\right) + \tan^{-1}\left(\tfrac{1}{3}\right)$ equals
A$\tfrac{\pi}{2}$
B$\tfrac{\pi}{3}$
C$\tfrac{\pi}{4}$
D$\pi$
Answer & Solution
Correct answer: C. $\tfrac{\pi}{4}$
Since $\tfrac{1}{2}\cdot\tfrac{1}{3} = \tfrac{1}{6} < 1$, apply $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\tfrac{x+y}{1-xy}$.
Here $\tfrac{x+y}{1-xy} = \tfrac{\tfrac{1}{2}+\tfrac{1}{3}}{1 - \tfrac{1}{6}} = \tfrac{5/6}{5/6} = 1$
so the sum is $\tan^{-1}(1) = \tfrac{\pi}{4}$.
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