If Tanπ4+θ+ Tanπ4-θ = a then Tan3π4+θ +Tan3π4-θ=

tanπ4+θ+ tanπ4-θ = a tan3π4+θ +tan3π4-θ =tanπ4+θ+tanπ4-θ3-tanπ4+θ tanπ4-θ tanπ4+θ+tanπ4-θ =a3-3(1)a =a3-3a

If Tanπ4+θ+ Tanπ4-θ = a then Tan3π4+θ +Tan3π4-θ=

tanπ4+θ+ tanπ4-θ = a tan3π4+θ +tan3π4-θ =tanπ4+θ+tanπ4-θ3-tanπ4+θ tanπ4-θ tanπ4+θ+tanπ4-θ =a3-3(1)a =a3-3a

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If $Tan (\frac{π}{4} + θ) + Tan (\frac{π}{4} - θ) = a$ then ${Tan}^{3} (\frac{π}{4} + θ) + {Tan}^{3} (\frac{π}{4} - θ) =$

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$a^{3} - 3 a$

answer is D.

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Detailed Solution

$\tan (\frac{π}{4} + θ) + \tan (\frac{π}{4} - θ) = a$

$\begin{array}{rcl} \tan^{3} (\frac{π}{4} + θ) + \tan^{3} (\frac{π}{4} - θ) \\ = & [\tan (\frac{π}{4} + θ) + \tan (\frac{π}{4} - θ)] 3 - \tan (\frac{π}{4} + θ) \\ \tan (\frac{π}{4} - θ) [\tan (\frac{π}{4} + θ) + \tan (\frac{π}{4} - θ)] \\ = & a^{3} - 3 (1) a \\ = & a^{3} - 3 a \end{array}$