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Prove that : $2tan^{-1} \bigg( \frac{1}{2} \bigg) + tan^{-1} \bigg( \frac{1}{7} \bigg) = tan^{-1} \bigg( \frac{31}{17} \bigg)$

Can you answer this question?

Toolbox:
• $2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}$
• $tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}, xy <1$
L.H.S $tan^{-1}\frac{4}{3}=tan^{-1}\frac{1}{7}$
$= tan^{-1}\frac{\frac{4}{3}=\frac{1}{7}}{1-\frac{4}{21}}=tan^{-1}\frac{31}{17}$ = R.H.S
answered Feb 28, 2013