# What is the derivative of sin x / cos ^2 x - 1 ?

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I come to the answer #(-cos x sin^2 x)/((cos^2x -1)^2)#

But it should be :

#(-cos x )/((cos^2x -1))#

I know #sin^2x = 1 - cos^2x#

But its not the same as #cos^2x -1#

So I do something wrong.

My first step was :

#((-2 cos x sinx sinx ) - (cos x (cos^2x -1)))/ (cos^2x -1)^2#

Simplify

#((-2cosx sin^2x) - cos x (cos^2x-1))/((cos^2x -1)^2)#

Gives:

#(-cosx((cos^2x-1)+2 sin^2x))/((cos^2x -1)^2)#

Gives:

#(-cosx(1+sin^2x-1))/((cos^2x -1)^2)#

Gives:

#(-cosxsin^2x)/((cos^2x -1)^2)#

I come to the answer

But it should be :

I know

But its not the same as

So I do something wrong.

My first step was :

Simplify

Gives:

Gives:

Gives:

##### 1 Answer

You're almost there!

What you have to realize is that

Once you understand that, it's just one step further to say that

Another error you have is that you reversed the order of the terms in the numerator when you did the quotient rule, remember,

This makes what you got off by a factor of

From *here*, use the identity

#=(cosxsin^2x)/(cos^2x-1)(-sin^2x))#

#=(-cosx)/(cos^2x-1)#

As you wished to show.

But, honestly, I think the *best* way to express this is by rewriting the remaining

#=(-cosx)/(-sin^2x)#

#=cosx/sinx(1/sinx)#

#=cotxcscx#

If you know your trigonometric derivatives, you might recognize that this is off the derivative of

So the derivative is exactly as we'd expect.