Prove that 2 tan ^{-1} frac{1}{2} + tan ^{-1} | vedclass

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$L$.$H$.$S$ $= 2 	an ^{-1} \frac{1}{2} + 	an ^{-1} \frac{1}{7}$ Using the formula $2 	an ^{-1} x = 	an ^{-1} \frac{2x}{1-x^2}$,we get: $= 	an ^{-

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$L$.$H$.$S$ $= 2 	an ^{-1} \frac{1}{2} + 	an ^{-1} \frac{1}{7}$ Using the formula $2 	an ^{-1} x = 	an ^{-1} \frac{2x}{1-x^2}$,we get: $= 	an ^{-1} \left( \frac{2 \cdot \frac{1}{2}}{1 - (\frac{1}{2})^2} ight) + 	an ^{-1} \frac{1}{7}$ $= 	an ^{-1} \left( \frac{1}{1 - \frac{1}{4}} ight) + 	an ^{-1} \frac{1}{7}$ $= 	an ^{-1} \left( \frac{1}{\frac{3}{4}} ight) + 	an ^{-1} \frac{1}{7} = 	an ^{-1} \frac{4}{3} + 	an ^{-1} \frac{1}{7}$ Now,using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$: $= 	an ^{-1} \left( \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \cdot \frac{1}{7}} ight)$ $= 	an ^{-1} \left( \frac{\frac{28+3}{21}}{\frac{21-4}{21}} ight) = 	an ^{-1} \left( \frac{31}{17} ight) = R.H.S$.

Prove that $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{31}{17}$.

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