Question
Prove that $\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=\pi$
$
\begin{aligned}
& \tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \text { if } x y>1 \\
& \text { Here } x y=1 \times 2=2>1 \\
\therefore \quad & \tan ^{-1} 1+\tan ^{-1} 2=\pi+\tan ^{-1}\left(\frac{1+2}{1-(1)(2)}\right) \\
= & \pi+\tan ^{-1}\left(\frac{3}{1-2}\right) \\
= & \pi+\tan ^{-1}(-3) \\
= & \pi-\tan ^{-1} 3\left(\text { As }, \tan ^{-1}(-x)=-\tan ^{-1} x\right) \\
\therefore & \tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=\pi
\end{aligned}
$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.