Question
Write the value of $\tan^{-1}\text{x}+\tan^{-1}\Big(\frac{1}{\text{x}}\Big)$ for x > 0.
✓
Answer
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big),\text{xy}<1$
$\therefore\ \tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}=\tan^{-1}\bigg(\frac{\text{x}+\frac{1}{\text{x}}}{1-\text{x}\frac{1}{\text{x}}}\bigg),\text{x}>0$
$=\tan^{-1}\Big(\frac{\text{x}^2+1}{0}\Big)$
$=\tan^{-1}(\infty)$
$=\tan^{-1}\Big(\tan\frac{\pi}{2}\Big)$
$=\frac{\pi}{2}$
$\therefore\ \tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{2}$
$\therefore\ \tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}=\tan^{-1}\bigg(\frac{\text{x}+\frac{1}{\text{x}}}{1-\text{x}\frac{1}{\text{x}}}\bigg),\text{x}>0$
$=\tan^{-1}\Big(\frac{\text{x}^2+1}{0}\Big)$
$=\tan^{-1}(\infty)$
$=\tan^{-1}\Big(\tan\frac{\pi}{2}\Big)$
$=\frac{\pi}{2}$
$\therefore\ \tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{2}$
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