allery$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $
A$\tfrac{\pi}{2}$
B$\tfrac{5\pi}{4}$
C$\tfrac{3\pi}{4}$
D$\pi$
Answer & Solution
Correct answer: D. $\pi$
For $\tan^{-1}(2) + \tan^{-1}(3)$, the product $2\cdot 3 = 6 > 1$
so use $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\tfrac{x+y}{1-xy}$ (positive case).
This gives $\pi + \tan^{-1}\tfrac{5}{-5} = \pi + \tan^{-1}(-1) = \pi - \tfrac{\pi}{4} = \tfrac{3\pi}{4}$.
Adding $\tan^{-1}(1) = \tfrac{\pi}{4}$: total $= \pi$.
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