Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) (cosec θ - cot θ)^2 = (1-cos θ)/(1+cos θ)

Dear Student

(cosec θ-cot θ)^2= (1-cos θ)/(1+cos θ)
To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)

L.H.S. = (cosec θ-cot θ)^2
The above equation is in the form of (a-b) , and expand it
Since (a-b)^2= a^2+ b^2–2ab
Here a = cosec θ and b = cot θ
= (cosec^2 θ +cot^2 θ-2cosec θ cot θ)
Now,
apply the corresponding inverse functions and equivalent ratios to simplify
= (1/sin^2 θ+ cos^2 θ/sin^2 θ-2cos θ/sin^2 θ)

= (1 + cos^2 θ-2cos θ)/(1 - cos^2 θ)

= (1-cos θ)^2 /(1 -cosθ)(1+cos θ)

= (1-cos θ)/(1+cos θ) =R.H.S.

Therefore,
(cosec θ-cot θ)^2= (1-cos θ)/(1+cos θ)
Hence proved.

Thanks