# Prove the trigonometric identity: (cot x)^2+1= (cosec x)^2.

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We have to prove that (cot x)^2 + 1 = (cosec x)^2

Let's start from the left hand side

(cot x)^2 + 1

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

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

=> [(cos x)^2 + (sin x)^2]/(sin x)^2

=> 1/(sin x)^2

=> (cosec x)^2

which is the right hand side

**This proves that: (cot x)^2 + 1 = (cosec x)^2**

We'll manage the right side, knowing that the cosecant function is given by the ratio:

`cosec x = 1/(sin x)`

We'll raise to square both sides:

`cosec^2 x = 1/(sin^2 x)`

We'll re-write the expression:

`cot^2 x + 1 = 1/(sin^2 x)`

We'll multiply both sides by `sin^2 x` :

`sin^2 x*cot^2 x + sin^2 x = 1`

But `cot^2 x = (cos^2 x)/(sin^2 x)`

The expression will become:

`(sin^2 x*cos^2x)/(sin^2 x) + sin^2 x = 1`

We'll simplify by `sin^2 x ` and we'll get the Pythagorean identity:

`cos^2 x + sin^2 x = 1`

`` **Since the given expression led to the Pythagorean identity, that means that the expression represents an identity, too.**