Browse Questions

# Show that $2tan^{-1}(-3)=\frac{-\pi}{2}+tan^{-1}\bigg(\frac{-4}{3}\bigg)$

Toolbox:
• $2tan^{-1}x=-\pi +tan^{-1}\large\frac{2x}{1-x^2} \: if \: x < -1$
• $tan^{-1}(-x)=-tan^{-1}x$
• $tan^{-1}x+tan^{-1}y=tan^{-1} \bigg(\large \frac{x+y}{1-xy} \bigg)\:\:if\:xy<1$
• $tan^{-1}\infty=\large\frac{\pi}{2}$
The question can be rewritten as prove $2tan^{-1}(-3)-tan^{-1} \bigg(\large\frac{-4}{3}\bigg)=-\large\frac{\pi}{2}$
By taking x=-3 we get $\large\frac{2x}{1-x^2}=\frac{2.-3}{1-9}=\large\frac{6}{8}=\large\frac{3}{4}$
Substituting in the above formula $2tan^{-1}(-3)=-\pi+tan^{-1}\large\frac{3}{4}$
$tan^{-1}(-\frac{4}{3})=-tan^{-1}\large\frac{4}{3}$

L.H.S.=
$2tan^{-1}(-3)-tan^{-1} \bigg(\large\frac{-4}{3}\bigg)$
$= -\pi+tan^{-1}\bigg( \large\frac{3}{4} \bigg)-tan^{-1} \bigg( \large\frac{-4}{3}\bigg)$
$= -\pi+tan^{-1}\large\frac{3}{4}+tan^{-1}\large\frac{4}{3}$
By taking $x=\large\frac{3}{4}\:and\:y=\large\frac{4}{3}$,
$\large\frac{x+y}{1-xy}=\large\frac{\large\frac{3}{4}+\large\frac{4}{3}}{1-\large\frac{3}{4}.\large\frac{4}{3}}=\large\frac{\large\frac{25}{12}}{1-1}=\infty$
$\Rightarrow tan^{-1}\large\frac{3}{4}+tan^{-1}\large\frac{4}{3}=tan^{-1}\infty$
$= -\pi+tan^{-1} \infty = -\pi+\large\frac{\pi}{2}$
$= \large\frac{-\pi}{2}$

edited Mar 19, 2013