If tan⁡π4+θ+tan⁡π4−θ=athen tan3⁡π4+θ+tan3⁡π4−θ=

tan3⁡π4+θ+tan3⁡π4−θ=a3−3(1)a=a3−3a using a3+b3=(a+b)3−3ab(a+b) ∵A+B=π2⇒tanAtanB=1

If tan⁡π4+θ+tan⁡π4−θ=athen tan3⁡π4+θ+tan3⁡π4−θ=

tan3⁡π4+θ+tan3⁡π4−θ=a3−3(1)a=a3−3a using a3+b3=(a+b)3−3ab(a+b) ∵A+B=π2⇒tanAtanB=1

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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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a³ – 3a

answer is D.

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$\begin{array}{l} \tan^{3} (\frac{π}{4} + θ) + \tan^{3} (\frac{π}{4} - θ) = a^{3} - 3 (1) a = a^{3} - 3 a \\ ({using a}^{3} + b^{3} = (a + b)^{3} - 3 a b (a + b)) (∵ A + B = \frac{π}{2} \Rightarrow \tan A \tan B = 1) \end{array}$

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If tan⁡π4+θ+tan⁡π4−θ=athen tan3⁡π4+θ+tan3⁡π4−θ=

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