Question

What is `cot^4theta/(1-cos^4theta)` in terms of non-exponential trigonometric functions?

Answer 1

`(cottheta*cottheta)/(sintheta*sintheta*sintheta*sintheta)`

Explanation:

This answer is incorrect!

`cot^4 theta = cos^4 theta/sin^4 theta`

`1-cos^2 theta = sin^2 theta`

Here is where the mistake was made:
`cot^4 theta / (1-cos^4 theta) = (cos^4 theta/sin^4 theta)/(sin^2 theta*cos^2 theta`

`(cos^4 theta/sin^4 theta)/(sin^2 theta*cos^2 theta) = (cos^2 theta/sin^4 theta)/(sin^2 theta)`

`(cos^2 theta/sin^4 theta)/(sin^2 theta/1)` or `cos^2 theta/sin^4 theta*1/sin^2 theta`

Multiply across

`cos^2 theta/sin^6 theta` = `cot^2 theta/sin^4 theta`

Expand to get rid of the exponents

`(cottheta*cottheta)/(sintheta*sintheta*sintheta*sintheta)`

I hope this is the form you were looking for!

Answer 2

`(cosx/sinx*cosx/sinx*cosx/sinx*cosx/sinx)/((sinx*sinx+sinx*sinx*cosx*cosx))`

Explanation:

`cot^4x/(1-cos^4x)=`
`cot^4x/((1-cos^2x)(1+cos^2x))=`
`cot^4x/((sin^2x)(1+cos^2x))=`

`(cosx/sinx*cosx/sinx*cosx/sinx*cosx/sinx)/((sinx*sinx+sinx*sinx*cosx*cosx))`