$\tan ^{-1} 2+\tan ^{-1} 3=$

$\tan ^{-1} 2+\tan ^{-1} 3=$
  1. $-\frac{\pi}{4}$
  2. $\frac{\pi}{4}$
  3. $\frac{3 \pi}{4}$
  4. $\frac{5 \pi}{4}$

Solution

$\begin{aligned} & \text { Since, } \tan ^{-1} 2+\tan ^{-1}(3) \\ = & \pi+\tan ^{-1}\left(\frac{2+3}{1-2 \times 3}\right)=\pi+\tan ^{-1}\left(\frac{5}{-5}\right)\end{aligned}$ $=\pi+\tan ^{-1}(-1)=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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