# Prove the following: $2 \: tan^{-1} \frac{1}{2} + tan ^{-1} \frac{1}{7} = tan^{-1} \frac{31}{17}$

Toolbox:
• $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2} for\:|x|<1$$
• $$tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\: \: \: xy<1$$
L.H.S:
$$\large\frac{1}{2}\:is\:<1\:so\:2tan^{-1}\large\frac{1}{2}=tan^{-1} \bigg( \large\frac{1}{1-\frac{1}{4}} \bigg)$$
=$$tan^{-1}\large\frac{4}{3}$$
$$2tan^{-1}\large\frac{1}{2}+tan^{-1}\large\frac{1}{7}=tan^{-1}\large\frac{4}{3}+tan^{-1}\large\frac{1}{7}$$

By taking $$x=\large\frac{4}{3}\:and\:y=\large\frac{1}{7}$$
$$\:and\:since \large\frac{4}{3}.\large\frac{1}{7}=\large\frac{4}{21}\:is\:<1$$ by using the above formula of $$tan^{-1}x+tan^{-1}y$$

$$\Rightarrow\:L.H.S.= tan^{-1} \bigg[ \large\frac{\large\frac{4}{3}+\large\frac{1}{7}}{1-\large\frac{4}{21}} \bigg]$$
$$= tan^{-1}\frac{31}{17}=R.H.S$$

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": 1, "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}