# Prove the identity cotx*sinx=cosx/(cos^2x+sin^2x)

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### 2 Answers

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

Now we know that (cos x)^2 + (sin x)^2 = 1

Also, cot x = cos x / sin x

So cot x*sin x = (cos x / sin x)* sin x = cos x

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

Therefore both the sides are equal to cos x.

**We prove that cot x*sin x = cos x /((cos x)^2 + (sin x)^2).**

We know that cot x = cos x/sin x

We also know that (sin x)^2 + (cos x)^2 = 1( the fundamental formula of trigonometry).

We'll re-write the identity:

(cos x/sin x)*sin x = cos x/1

We'll simplify by sin x, to the left side and we'll get:

cos x = cos x, true

**The identity is proven and cotx*sinx=cosx/(cos^2x+sin^2x), for any measure of the angle x.**