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?

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.