Question

(1) If sin⁡θ=cot⁡α  tan⁡γ and tan⁡θ=cosα tanβ then prove that one of the value of cos⁡θ=cosβsecγ

(2) If sin²θ = cos³θ, prove that, cot^6θ - cot^2θ = 1​

Answer 1

Answer:

Step-by-step explanation:

(1) sin⁡θ=cot⁡α tan⁡γ

    tan⁡θ=cosα tanβ

    => Sinθ/Cosθ = cosα tanβ

     => Cotαtanγ/Cosαtanβ = Cosθ

     => tanγ/Sinαtanβ = Cosθ

     => Sinγ/Cosγ / Sinα.Sinβ/Cosβ = Cosθ

     => Cosθ = Cosβ * Sinγ /Cosγ * Sinα * Sinβ

     => Cosθ =  (CosβSecγ) (Sinγ/SinαSinβ)

(2) Sin²θ = Cos³θ

   Cot⁶θ - Cot²θ  

=> Cot⁶θ - Cot²θ  

=> Cos⁶θ/Sin⁶θ  - Cos²θ / Sin²θ

=> (Cos³θ)² / Sin⁶θ - Cos²θ/Cos³θ

=> Sin⁴θ/Sin⁶θ - 1/Cosθ    (∵ Cos³θ = Sin²θ)

=> 1/Sin²θ - 1/Cosθ

=> 1/Cos³θ - 1/Cosθ     (∵ Sin²θ = Cos³θ)

=> 1 - Cos²θ / Cos³θ

=> Sin²θ / Sin²θ    (∵ 1 - Cos²θ = Sin²θ)

=> 1

= R.H.S

Hence proved.

Answer 2

(1) Please look at the picture attached. sin²theta should be written in the whitenered area for this question.

(2) cot^6x - cot²x = cos^6x/sin^6x - cos²x/sin²x = (cos^3x)²/sin^6x - cos²x/cos³x = (sin²x)²/sin^6x - 1/cosx = 1/sin²x - 1/cos x = 1/cos³x - 1/cosx = sin²x/cos³x = sin²x/sin²x = 1