# What does $\tan^{-1}{\frac{1}{4}}+\tan^{-1}{\frac{2}{9}}$ reduce to?

$\begin{array}{1 1} sin^{-1}\frac{1}{\sqrt5} \\ sin^{-1}\frac{2}{\sqrt5} \\ cos^{-1}\frac{1}{\sqrt5} \\ tan^{-1}\frac{1}{\sqrt5} \end{array}$

Toolbox:
• $$tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}\:\:\:xy<1$$
• $$tan^{-1}x=sin^{-1}\frac{x}{\sqrt{1+x^2}}$$
By taking $$x=\frac{1}{4}\:and\:y=\frac{2}{9}\:we\:get$$
$$\frac{x+y}{1-xy}=\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}.\frac{2}{9}}=\frac{17}{36}.\frac{36}{34}=\frac{17}{34}=\frac{1}{2}$$
Substituting in the above formula we get L.H.S.=
$$tan^{-1}\frac{1}{4}+tan^{-1}\frac{2}{9}=tan^{-1}\frac{1}{2}$$
By taking $$x=\frac{1}{2},\:\frac{x}{\sqrt{1+x^2}}=\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{4}}}=\frac{1}{2}.\frac{2}{\sqrt{5}}=\frac{1}{\sqrt5}$$
Substituting in the above formula of $$tan^{-1}x$$ we get
$$tan^{-1}\frac{1}{2}= sin^{-1}\frac{1}{\sqrt 5}$$
edited Jul 9, 2014