tanθ+tan2θ=tan3θ ((tanθ+tan2θ)/(1−tanθtan2θ))(1−tanθtan2θ)=tan3θ tan3θ(1−tanθtan2θ)=tan3θ tanθtan2θtan3θ 0 3θ=kπ, where k=0,1,2,3,... θ=((kπ)/3) tanθ=0 or tan2θ=0 or tan 3θ = 0 θ=mπ, where m=0,1,2,3,... 2θ=nπ, where n=0,1,2,3,... θ=((nπ)/2) but θ∉(((p+1)π)/2), p=0,1,2,3,... The exhaustive set ={θ/θ=((kπ)/3) or θ=mπ}


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Question Number 14330 by myintkhaing last updated on 01/Jun/17

$t a n θ + t a n 2 θ = t a n 3 θ$ $\frac{t a n θ + t a n 2 θ}{1 - t a n θ t a n 2 θ} (1 - t a n θ t a n 2 θ) = t a n 3 θ$ $t a n 3 θ (1 - t a n θ t a n 2 θ) = t a n 3 θ$ $t a n θ t a n 2 θ t a n 3 θ 0$ $3 θ = k π, w h e r e k = 0, 1, 2, 3, . . .$ $θ = \frac{k π}{3}$ $t a n θ = 0 o r t a n 2 θ = 0 o r t a n 3 θ = 0$ $θ = m π, w h e r e m = 0, 1, 2, 3, . . .$ $2 θ = n π, w h e r e n = 0, 1, 2, 3, . . .$ $θ = \frac{n π}{2} b u t θ \notin \frac{(p + 1) π}{2}, p = 0, 1, 2, 3, . . .$ $T h e e x h a u s t i v e s e t = {θ / θ = \frac{k π}{3} o r θ = m π}$
Commented by Tinkutara last updated on 30/May/17

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