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

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(A) We use the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$.Given $L.H.S. = 	an ^{-1} \frac{1}{2} + 	an ^{-1}

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(A) We use the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$.Given $L.H.S. = 	an ^{-1} \frac{1}{2} + 	an ^{-1} \frac{2}{11}$.Applying the formula with $x = \frac{1}{2}$ and $y = \frac{2}{11}$:$L.H.S. = 	an ^{-1} \left( \frac{\frac{1}{2} + \frac{2}{11}}{1 - (\frac{1}{2} 	imes \frac{2}{11})} ight)$.Simplifying the numerator: $\frac{1}{2} + \frac{2}{11} = \frac{11 + 4}{22} = \frac{15}{22}$.Simplifying the denominator: $1 - \frac{2}{22} = 1 - \frac{1}{11} = \frac{10}{11}$.Thus,$L.H.S. = 	an ^{-1} \left( \frac{15/22}{10/11} ight) = 	an ^{-1} \left( \frac{15}{22} 	imes \frac{11}{10} ight)$.$L.H.S. = 	an ^{-1} \left( \frac{15}{2 	imes 10} ight) = 	an ^{-1} \left( \frac{15}{20} ight) = 	an ^{-1} \frac{3}{4}$.Since $L.H.S. = R.H.S.$,the identity is proved.

Show that $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} \frac{3}{4}$

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