Prove tan−121+tan−1112=tan−143

LHS=tan−121+tan−1112=tan−1[(1−(21)(112)21+112)]=tan−1((2222−2)(2211+4))=tan−12015=tan−143=RHS

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Prove ^ { - 1 } { 1 } { 2 } + ^ { - 1 } { 2 } { 11 } = ^

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

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

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$L H S = tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{2}{11} = tan^{- 1} [(\frac{\frac{1}{2} + \frac{2}{11}}{1 - ( \frac{1}{2} ) ( \frac{2}{11} )})] = tan^{- 1} (\frac{( \frac{11 + 4}{22} )}{( \frac{22 - 2}{22} )}) = tan^{- 1} \frac{15}{20} = tan^{- 1} \frac{3}{4} = R H S$