Discussion
The value of \(2\tan^{-1}\left(\frac{1}{3}\right) + \cot^{-1}\left(\frac{3}{4}\right) =\)
Answer: E
The value is \(\frac{\pi}{2}\).
Use \(2\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^{2}}\) for \(|x| \lt 1\):
\[ \begin{aligned} 2\tan^{-1}\frac{1}{3} &= \tan^{-1}\frac{2/3}{8/9} = \tan^{-1}\frac{3}{4} \\ \cot^{-1}\frac{3}{4} &= \tan^{-1}\frac{4}{3} \end{aligned} \]
Then \(\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{4}{3} = \frac{\pi}{2}\), because \(\tan^{-1}y + \tan^{-1}\frac{1}{y} = \frac{\pi}{2}\) for \(y \gt 0\).
The common slip is to treat \(\cot^{-1}\frac{3}{4}\) as \(\tan^{-1}\frac{3}{4}\); it is \(\tan^{-1}\frac{4}{3}\), the complementary angle, which is what makes the sum exactly a right angle.
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