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

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A)
• $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2} for\:|x|<1$$
• $$tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\: \: \: xy<1$$
$$\large\frac{1}{2}\:is\:<1\:so\:2tan^{-1}\large\frac{1}{2}=tan^{-1} \bigg( \large\frac{1}{1-\frac{1}{4}} \bigg)$$
=$$tan^{-1}\large\frac{4}{3}$$
$$2tan^{-1}\large\frac{1}{2}+tan^{-1}\large\frac{1}{7}=tan^{-1}\large\frac{4}{3}+tan^{-1}\large\frac{1}{7}$$
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 $$tan^{-1}x+tan^{-1}y$$
$$\Rightarrow\:L.H.S.= tan^{-1} \bigg[ \large\frac{\large\frac{4}{3}+\large\frac{1}{7}}{1-\large\frac{4}{21}} \bigg]$$
$$= tan^{-1}\frac{31}{17}=R.H.S$$