Question
Solve:
$5\tan^{-1}\text{x}+3\cot^{-1}\text{x}={2\pi}$
$5\tan^{-1}\text{x}+3\cot^{-1}\text{x}={2\pi}$
✓
Answer
$5\tan^{-1}\text{x}+3\cot^{-1}\text{x}={2\pi}$
$\Rightarrow5\tan^{-1}\text{x}+3\Big(\frac{\pi}{2}-\tan^{-1}\text{x}\Big)={2\pi}$
$\Big[\because\ \cot^{-1}\text{x}=\frac{\pi}{2}-\tan^{-1}\text{x}\Big]$
$\Rightarrow5\tan^{-1}\text{x}+\frac{3\pi}{2}-3\tan^{-1}\text{x}={2\pi}$
$\Rightarrow2\tan^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{4}$
$\Rightarrow\text{x}=\tan\frac{\pi}{4}=1$
$\Rightarrow5\tan^{-1}\text{x}+3\Big(\frac{\pi}{2}-\tan^{-1}\text{x}\Big)={2\pi}$
$\Big[\because\ \cot^{-1}\text{x}=\frac{\pi}{2}-\tan^{-1}\text{x}\Big]$
$\Rightarrow5\tan^{-1}\text{x}+\frac{3\pi}{2}-3\tan^{-1}\text{x}={2\pi}$
$\Rightarrow2\tan^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{4}$
$\Rightarrow\text{x}=\tan\frac{\pi}{4}=1$
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