# Prove that $tan^{-1} \frac{1}{5}+ tan^{-1} \frac{1}{7}+tan^{-1}\frac{1}{3}+tan^{-1} \frac{1}{8}= \frac{\pi}{4}$

Toolbox:
• $$tan^{-1}x=tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg) xy < 1$$
Given $tan^{-1} \large\frac{1}{5}+ tan^{-1} \large\frac{1}{7}+tan^{-1}\large\frac{1}{3}+tan^{-1} \large\frac{1}{8}= \large\frac{\pi}{4}$

We will group two of the terms in the above L.H.S. equation and solve them using known inverse trignometric identities.
$$\Rightarrow L.H.S. = \bigg( tan^{-1}\large\frac{1}{5}+tan^{-1}\large\frac{1}{7}\bigg) + \bigg( tan^{-1}\large\frac{1}{3}+tan^{-1}\large\frac{1}{8}\bigg)$$

Let's take the first term $tan^{-1}\large\frac{1}{5}+tan^{-1}\large\frac{1}{7}$
We know that $$tan^{-1}x=tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg) xy < 1$$

By taking $$x=\large\frac{1}{5}\:and\:y=\large\frac{1}{7}$$ in the above formula,
$x+y = \large\frac{1}{5}+\large\frac{1}{7} = \large\frac{12}{35}$
$1 - xy = 1 -\large \frac{1}{5} \times \frac{1}{7} = \large\frac{34}{35}$
$\large \frac{x+y}{1-xy}$ $= \large\frac{12}{35}.\large\frac{35}{34}=\large\frac{12}{34} = \large\frac{6}{17}$
$$\Rightarrow tan^{-1}\large\frac{1}{5}+tan^{-1}\large\frac{1}{7}=tan^{-1} \large\frac{6}{17}$$

Let's take the second term: $tan^{-1}\large\frac{1}{3}+tan^{-1}\large\frac{1}{8}$

Similarly by taking $$x=\large\frac{1}{3}\:and\:y=\large\frac{1}{8}$$ we get:
$x+y = \large\frac{1}{3}+\large\frac{1}{8} = \large\frac{11}{24}$
$1-xy = 1-\large\frac{1}{3}\times \frac{1}{8} = \large\frac{23}{24}$
$$\large \frac{x+y}{1-xy}$$ $=\large\frac{11}{24}.\large\frac{24}{23}=\large\frac{11}{23}$
$\Rightarrow tan^{-1}\large\frac{1}{3}+tan^{-1}\large\frac{1}{8} = tan^{-1}\large\frac{11}{23}$

Therefore $tan^{-1} \large\frac{1}{5}+ tan^{-1} \large\frac{1}{7}+tan^{-1}\large\frac{1}{3}+tan^{-1}\large \frac{1}{8}$ now reduces to $$\Rightarrow\: tan^{-1} \large\frac{6}{17}+tan^{-1}\large\frac{11}{23}$$

Similarly by taking $$x=\large\frac{6}{17}\:and\:y=\large\frac{11}{23}$$ we get:
$x+y = \large\frac{6}{17}+\large\frac{11}{23} = \large\frac{138+187}{391} =\large\frac{325}{391}$
$1-xy = 1 -\large \frac{6}{17} \times \frac{11}{23} = 1-\large\frac{66}{391} = =\large\frac{325}{391}$
$$\large \frac{x+y}{1-xy}=\large \frac{325}{391}\times \frac{391}{325}=1$$

$$\Rightarrow$$ L.H.S. $$= tan^{-1} \large\frac{6}{17}+tan^{-1}\large\frac{11}{23} = tan^{-1}1$$ $$= \large\frac{\pi}{4}$$ = R.H.S.

edited Mar 15, 2013
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}