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If $\tan^{-1}\left(\dfrac{1}{2}\right) + \tan^{-1}\left(\dfrac{1}{3}\right) = ?$
A$\dfrac{\pi}{3}$
B$\dfrac{\pi}{2}$
C$\dfrac{\pi}{6}$
D$\dfrac{\pi}{4}$
Answer & Solution
Correct answer: D. $\dfrac{\pi}{4}$
Use the identity $\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\dfrac{a + b}{1 - ab}\right)$, valid when $ab < 1$.
With $a = 1/2$, $b = 1/3$: $ab = 1/6 < 1$, so the identity applies cleanly.
$\dfrac{a + b}{1 - ab} = \dfrac{1/2 + 1/3}{1 - 1/6} = \dfrac{5/6}{5/6} = 1$.
So the sum equals $\tan^{-1}(1) = \dfrac{\pi}{4}$.
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