Q184 / 270 · NDA 2011 (Phase 2) · Mathematics

In a triangle $A B C$, if $A=\tan ^{-1} 2$ and $B=\tan ^{-1} 3$, then $C$ is equal to

  • A
    $\frac{\pi}{3}$
  • Correct answer
    B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{6}$
  • D
    $\frac{\pi}{2}$

Solution

$\begin{aligned} & \tan \mathrm{A}=2 \text { and } \tan \mathrm{B}=3 \\ & \qquad \begin{array}{c}\tan (\mathrm{A}+\mathrm{B})= \\ =\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A} \cdot \tan \mathrm{B}}=\frac{2+3}{1-6} \\ =\frac{5}{-5}=-1 \\ \because \quad \mathrm{~A}+\mathrm{B}+\mathrm{C}=\pi \\ \mathrm{A}+\mathrm{B}=\pi-\mathrm{C}\end{array} \\ & \tan (\mathrm{A}+\mathrm{B})=\tan (\pi-\mathrm{C})=-\tan \mathrm{C} \\ & \therefore-\tan \mathrm{C}=-1 \Rightarrow \tan \mathrm{C}=\tan \frac{\pi}{4} \\ & \therefore \mathrm{C}=\frac{\pi}{4} .\end{aligned}$

Hence, the correct answer is option: 'B'.

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