# Trigonometry problem.Prove that (cot x)'=-csc^2 x

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### 2 Answers

You also may use the following trigonometric identity such that:

`cot x = 1/tan x`

Differentiating with respect to x yields:

`(cot x)' = (1/tan x)' `

You need to differentiate using derivatives rules, such that:

`(1/tan x)'= (1'*tan x - 1*(tan x)')/(tan^2 x)`

`(1/tan x)'= (0*tan x - 1*(1/cos^2 x))/(tan^2 x)`

Replacing `sin^2 x/cos^2 x` for `tan^2 x` yields:

`(1/tan x)'= (- 1*(1/cos^2 x))/(sin^2 x/cos^2 x)`

Reducing suplicate factors yields:

`(1/tan x)'= (-1/sin^2 x) => (cot x)' = (-1/sin^2 x) `

Since `1/sin x = csc x => 1/sin^2 x = csc^2 x` , such that:

`(cot x)' = -csc^2 x`

**Hence, testing if `(cot x)' = - csc^2 x` using the rules of differentiation and trigonometric identities yields that `(cot x)' = - csc^2 x` holds.**

We know that cosec x = 1/sin x and cot x = cos x/ sin x

If we have to calculate the derivative of cot x, we'll consider the quotient rule:

(cos x/sin x)' = (cos x)'*(sin x) - (cos x)*(sin x)'/(sin x)^2

We'llĀ calculate the derivatives of the functions sine and cosine:

(cos x)' = -sin x

(sin x)' = cos x

(cot x)' = [(-sin x)*(sin x) - (cos x)^2]/(sin x)^2

(cot x)' = - [(sin x)^2 + (cos x)^2]/(sin x)^2

But, from the fundamental formula of trigonometry, we'll have:

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

(cot x)' = -1/(sin x)^2

But 1/sin x = csc x.

If we'll raise to square both sides, we'll get:

1/(sin x)^2 = -(csc x)^2