Question

$If \\ \\ tan \ \theta + cot \ \theta = 3, \\ \\ then, \\ \\ tan^8 \ \theta + cot^8 \ \theta = \ ?$

Answer 1

Given : Tanθ + Cotθ = 3

Squaring on both Sides, We get :

⇒ (Tanθ + Cotθ)² = 3²

⇒ Tan²θ + Cot²θ + 2Tanθ × Cotθ = 9

We know that : Tanθ × Cotθ = 1

⇒ Tan²θ + Cot²θ + 2 = 9

⇒ Tan²θ + Cot²θ  = 7

Again Squaring on both Sides, We get :

⇒ (Tan²θ + Cot²θ)² = 7²

⇒ Tan⁴θ + Cot⁴θ + 2(Tanθ × Cotθ)² = 49

We know that : Tanθ × Cotθ = 1

⇒ Tan⁴θ + Cot⁴θ + 2 = 49

⇒ Tan⁴θ + Cot⁴θ  = 47

Again Squaring on both Sides, We get :

⇒ (Tan⁴θ + Cot⁴θ)² = 47²

⇒ Tan⁸θ + Cot⁸θ + 2(Tanθ × Cotθ)⁴ = 2209

We know that : Tanθ × Cotθ = 1

⇒ Tan⁸θ + Cot⁸θ + 2 = 2209

⇒ Tan⁸θ + Cot⁸θ  = 2207

Answer 2

hey buddy refer attachment..


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