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