# 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