If Tan-1tan5π4=α and Tan-1-tan2π3=β then

α=tan-1tanπ+π4=tan-1tan π4=π4 β=tan-1-tanπ-π3=tan-1tan π3=π3 α+β=π4+π3=7π12

If Tan-1tan5π4=α and Tan-1-tan2π3=β then

α=tan-1tanπ+π4=tan-1tan π4=π4 β=tan-1-tanπ-π3=tan-1tan π3=π3 α+β=π4+π3=7π12

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If ${Tan}^{- 1} (\tan \frac{5 π}{4}) = α$ and ${Tan}^{- 1} (- \tan \frac{2 π}{3}) = β$ then

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$α - β = \frac{7 π}{12}$

$α + β = \frac{7 π}{12}$

$2 α + 3 β = \frac{7 π}{12}$

$4 α + 3 β = \frac{7 π}{12}$

answer is B.

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Detailed Solution

$α = \tan^{- 1} (\tan (π + \frac{π}{4})) = \tan^{- 1} (\tan \frac{π}{4}) = \frac{π}{4} β = \tan^{- 1} (- \tan (π - \frac{π}{3})) = \tan^{- 1} (\tan \frac{π}{3}) = \frac{π}{3} α + β = \frac{π}{4} + \frac{π}{3} = \frac{7 π}{12}$

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If Tan-1tan5π4=α and Tan-1-tan2π3=β then

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If ${Tan}^{- 1} (\tan \frac{5 π}{4}) = α$ and ${Tan}^{- 1} (- \tan \frac{2 π}{3}) = β$ then