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

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Answer 1 · 2018-03-14T02:22:21

$(cottheta*cottheta)/(sintheta*sintheta*sintheta*sintheta)$

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 · 2018-03-14T02:57:45

$(cosx/sinx*cosx/sinx*cosx/sinx*cosx/sinx)/((sinx*sinx+sinx*sinx*cosx*cosx))$

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

$(cosx/sinx*cosx/sinx*cosx/sinx*cosx/sinx)/((sinx*sinx+sinx*sinx*cosx*cosx))$

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