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Q.

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

3a

b

a

c

0

d

$a^{3} - 3 a$

answer is D.

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Detailed Solution

$\tan^{3} (\frac{π}{4} + θ) + \tan^{3} (\frac{π}{4} - θ)$ $= a^{3} - 3 (1) a = a^{3} - 3 a$

$(u sing A^{3} + B^{3} = (A + B)^{3} - 3 A B (A + B))$

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