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I recently started learning the inverse trigonometric function and got stuck at one point.

In the question involving the expression of tangents, It was given that $\frac{x}{y}>1$

The authors opened $$\tan^{-1}\frac {1+{x \over y} }{1-{x \over y}}$$ as $$-\pi + \tan^{-1}1+\tan^{-1} x/y$$ by giving the reason that ${x\over y} >1$

My doubt is, that I know that $\tan x \in [0,\infty)$ in the first quadrant. So, why did the author took $-\pi+\tan^{-1}1+\tan^{-1} x/y$ when we could have found that angle concerning $x$ in the first quadrant also?

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    $\begingroup$ Notice that since $x/y>1$ the expression in the arctangent is negative. By convention, $\tan^{-1}\theta$ is between $-\pi$ and $\pi$ so the answer you are looking for has to be in $(-\pi, 0)$.. $\endgroup$ – Mark Fischler Jun 23 '16 at 19:03
  • $\begingroup$ @MarkFischler Thanks. $\endgroup$ – Harsh Sharma Jun 23 '16 at 19:04
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As $\dfrac{x}{y} > 1$, so $\dfrac{1+\frac{x}{y}}{1-\frac{x}{y}} < 0$

According to you $\dfrac{1+\frac{x}{y}}{1-\frac{x}{y}} =\tan^{-1}{1}+\tan^{-1}{\frac{x}{y}}$, which is wrong as $\tan^{-1}{1}=\frac{\pi}{4}$, so $\tan^{-1}{1}+\tan^{-1}{\frac{x}{y}}$ will be in the second quadrant.

This is incorrect, since the principal range of $\tan^{-1}$ is from $-\pi/2$ to $\pi/2$, so we add a $(-\pi)$ in the beginning.

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