# Prove the identity: sin^2x + 2cos^2x = csc^2x - cos^2x*cot^2x

justaguide | Certified Educator

We have to prove that (sin x)^2 + (cos x)^2 = (csc x)^2 - (cos x)^2*(cot x)^2

Take the right hand side

(csc x)^2 - (cos x)^2*(cot x)^2

csc x = 1/ sin x and cot x = (cos x)/(sin x)

=> 1/(sin x)^2 - (cos x)^2*(cos x)^2/(sin x)^2

=> (1 - (cos x)^4)/(sin x)^2

=> (1 - (cos x)^2)(1 + (cos x)^2)/(sin x)^2

=> (sin x)^2(1 + (cos x)^2)/(sin x)^2

=> 1 + (cos x)^2

=> (sin x)^2 + (cos x)^2 + (cos x)^2

=> (sin x)^2 + 2*(cos x)^2

which is the left hand side

This proves the given identity.

giorgiana1976 | Student

The identity that has to be verified is

(sin x)^2 + 2(cos x)^2 = (csc x)^2 - (cos x)^2(cot x)^2

First, we'll manage the left side and we'll write

(sinx)^2 + 2(cos x)^2 = (sinx)^2 + (cos x)^2 + (cos x)^2

We'll use the Pythagorean identity:

(sinx)^2 + (cos x)^2 = 1

(sinx)^2 + 2(cos x)^2 = 1 + (cos x)^2

We'll manage the right side and we'll write:

(csc x)^2 - (cos x)^2(cot x)^2 = 1/(sinx)^2 - (cos x)^2*(cos x)^2/(sinx)^2

(csc x)^2 - (cos x)^2(cot x)^2 = [1-(cos x)^4]/(sinx)^2

We'll have a difference of two squares that returns the product:

(csc x)^2 - (cos x)^2(cot x)^2 = [1 + (cos x)^2][1 - (cos x)^2]/(sinx)^2

But 1 - (cos x)^2= (sinx)^2

(csc x)^2 - (cos x)^2(cot x)^2 = [1 + (cos x)^2]*(sinx)^2/(sinx)^2

We'll simplify and we'll get:

(csc x)^2 - (cos x)^2(cot x)^2 = [1 + (cos x)^2]

We notice that managing both sides, we'll get the same result [1 + (cos x)^2], therefore the identity (sin x)^2 + 2(cos x)^2 = (csc x)^2 - (cos x)^2(cot x)^2 is verified.