Question

Prove that: `(cos theta - 2 cos^3 theta)/(sin theta - 2 sin^3 theta) + cot theta = 0`.

Answer 1

Given:

`(cos theta - 2 cos^3 theta)/(sin theta - 2 sin^3 theta) + cot theta = 0`

Step 1: Factorization of numerator and denominator

`cos^3 theta = cos theta · cos^2 theta`

`sin^3 theta = sin theta · sin^2 theta`

Now, rewrite the given fraction: 

`(cos theta - 2 cos^3 theta)/(sin theta - 2 sin^3 theta)`

Factor out cos⁡ θ and sin θ:

`(cos theta(1 - 2cos^2 theta))/(sin theta (1 - 2sin^2 theta))`

1 = 2sin²θ = cos2θ

Fraction simplifies to:

`(cos theta (- cos2theta))/(sin theta(cos 2theta))`

`(-cos theta)/(sin theta)`

Using the identity `(-cos theta)/(sin theta) = cot theta`, we get 

−cot θ

Thus, adding cot⁡ θ:

−cot θ + cotθ = 0

`(cos theta - 2 cos^3 theta)/(sin theta - 2 sin^3 theta) + cot theta = 0`

Hence proved.