# Find the value of the following function: $tan^{-1} \frac{1}{5}+ tan^{-1} \frac{1}{7}+tan^{-1}\frac{1}{3}+tan^{-1} \frac{1}{8}$

Toolbox:
• $$tan^{-1}x=tan^{-1}y=tan^{-1} \bigg( \frac{x+y}{1-xy} \bigg) xy < 1$$
Given $tan^{-1} \frac{1}{5}+ tan^{-1} \frac{1}{7}+tan^{-1}\frac{1}{3}+tan^{-1} \frac{1}{8}$
We will group two of the terms in the above L.H.S. equation and solve them using known inverse trignometric identities.
$$\Rightarrow L.H.S. = \bigg( tan^{-1}\frac{1}{5}+tan^{-1}\frac{1}{7}\bigg) + \bigg( tan^{-1}\frac{1}{3}+tan^{-1}\frac{1}{8}\bigg)$$
Let's take the first term $tan^{-1}\frac{1}{5}+tan^{-1}\frac{1}{7}$
We know that $$tan^{-1}x=tan^{-1}y=tan^{-1} \bigg( \frac{x+y}{1-xy} \bigg) xy < 1$$
By taking $$x=\frac{1}{5}\:and\:y=\frac{1}{7}$$ in the above formula,
$x+y = \frac{1}{5}+\frac{1}{7} = \frac{12}{35}$
$1 - xy = 1 - \frac{1}{5} \times \frac{1}{7} = \frac{34}{35}$
$\large \frac{x+y}{1-xy}$ $= \frac{12}{35}.\frac{35}{34}=\frac{12}{34} = \frac{6}{17}$
$$\Rightarrow tan^{-1}\frac{1}{5}+tan^{-1}\frac{1}{7}=tan^{-1} \frac{6}{17}$$
Let's take the second term: $tan^{-1}\frac{1}{3}+tan^{-1}\frac{1}{8}$
Similarly by taking $$x=\frac{1}{3}\:and\:y=\frac{1}{8}$$ we get:
$x+y = \frac{1}{3}+\frac{1}{8} = \frac{11}{24}$
$1-xy = 1-\frac{1}{3}\times \frac{1}{8} = \frac{23}{24}$
$$\large \frac{x+y}{1-xy}$$ $=\frac{11}{24}.\frac{24}{23}=\frac{11}{23}$
$\Rightarrow tan^{-1}\frac{1}{3}+tan^{-1}\frac{1}{8} = tan^{-1}\frac{11}{23}$
Therefore $tan^{-1} \frac{1}{5}+ tan^{-1} \frac{1}{7}+tan^{-1}\frac{1}{3}+tan^{-1} \frac{1}{8}$ now reduces to $$\Rightarrow\: tan^{-1} \frac{6}{17}+tan^{-1}\frac{11}{23}$$
Similarly by taking $$x=\frac{6}{17}\:and\:y=\frac{11}{23}$$ we get:
$x+y = \frac{6}{17}+\frac{11}{23} = \frac{138+187}{391} =\frac{325}{391}$
$1-xy = 1 - \frac{6}{17} \times \frac{11}{23} = 1-\frac{66}{391} = =\frac{325}{391}$
$$\large \frac{x+y}{1-xy}= \frac{325}{391}\times \frac{391}{325}=1$$
$$\Rightarrow$$ $$= tan^{-1} \frac{6}{17}+tan^{-1}\frac{11}{23} = tan^{-1}1$$ $$= \large \frac{\pi}{4}$$
answered Mar 14, 2013