If tanθ + cotθ = 2, then tan³θ + cot³θ equals:

2. Since tanθ·cotθ = 1, use a³ + b³ = (a + b)³ − 3ab(a + b) = 2³ − 3(1)(2) = 8 − 6 = 2.

If tanθ + cotθ = 2, then ta $n^{3}$ θ + co $t^{3}$ θ equals:

Competition practiceTrigonometry

Since tanθ·cotθ = 1, use

a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b) = 2^{3} - 3 (1) (2) = 8 - 6 = 2

Answer: D) 2

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