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

2 Answers
Mar 14, 2018

#(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!

Mar 14, 2018

#(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))#