Question

Solve: $ta{n}^{-1}1+ta{n}^{-1}2+ta{n}^{-1}3=\pi$

Answer 1

LHS:
${\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3$
$={\tan }^{-1}1+\frac{\pi }{2}-{\cot }^{-1}2+\frac{\pi }{2}-{\cot }^{-1}3$ $(\because {\tan }^{-1}x=\frac{\pi }{2}-{\cot }^{-1}x$ $\forall x\in R)$
$=\pi +{\tan }^{-1}1-{\tan }^{-1}\frac{1}{2}-{\tan }^{-1}\frac{1}{3}$ $(\because {\tan }^{-1}x={\cot }^{-1}\frac{1}{x}$ $\forall x\in R^{+})$
$=\pi +{\tan }^{-1}1-{\tan }^{-1}(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot \frac{1}{3}})$ $(\because {\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}(\frac{a+b}{1-ab})$ if $ab<1$)
$=\pi +{\tan }^{-1}1-{\tan }^{-1}(\frac{5/6}{5/6})$
$=\pi +{\tan }^{-1}1-{\tan }^{-1}1$
$=\pi$ = RHS
Hence, proved.