How do you express #cot^3theta+tan^3theta # in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-10-15T11:03:02

#cot^3theta+tan^3theta #

#=cos^3theta/sin^3theta+sin^3theta/cos^3theta#

#=(cos^6theta+sin^6theta)/(sin^3thetacos^3theta)#

#=((cos^2theta+sin^2theta)^3-3(cos^2thetasin^2theta)(cos^2theta+sin^2theta))/(sin^3thetacos^3theta)#

#=(1-3(cos^2thetasin^2theta)*1)/(sin^3thetacos^3theta)#

#=(8-24(cos^2thetasin^2theta))/(2^3sin^3thetacos^3theta)#

#=(8-6(2costhetasintheta)^2)/(2sinthetacostheta)^3#

#=(8-3*2sin^2 2theta)/(sin^3 2theta)#

#=(8-3*(1-cos4theta))/(1/4(3sin2theta-sin6theta)#

#=(5+3cos4theta)/(1/4(3sin2theta-sin6theta)#

#=(20+12cos4theta)/(3sin2theta-sin6theta)#