# Prove that (sin x)^2*[1 + (cot x)^2] = 1

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### 2 Answers

We have to prove that (sin x)^2 * [ 1 + (cot x)^2] = 1

(sin x)^2 * [ 1 + (cot 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 + (cos x)^2

=> 1

**This proves that (sin x)^2 * [ 1 + (cot x)^2] = 1**

We know, as a consequence of Pythagorean identity, that:

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

Let's see how:

Pythagorean identity states that:

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

We'll divide by (sin x)^2:

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

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

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

Now, we'll substitute what's inside brackets by the equivalent above:

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