Prove the identity: sin^2x + 2cos^2x = csc^2x

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.