How do you express #cot^3theta-cos^2theta-tan^2theta # in terms of non-exponential trigonometric functions?

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Answer 1 · 2018-06-09T15:49:32

#(cos^5(x)-sin^5(x)-cos^4(x)sin^3(x))/(sin^3(x)*cos^2(x))#

We write
#1/tan^3(x)-tan^2(x)-cos^2(x)#

#(1-tan^5(x))/tan^3(x)-cos^2(x)#

#(1-tan^5(x)-tan^3(x)*cos^2(x))/tan^3(x)#

#(1-tan^5(x)-sin^2(x)/cos(x))/tan^3(x)#

#(cos(x)-cos(x)tan^5(x)-sin^3(x))/(cos(x)tan^3(x))#

#(cos^5(x)-sin^5(x)-cos^4(x)sin^3(x))/(sin^3(x)cos^2(x))#

Answer 2 · 2018-08-10T00:31:26

#( cos 3theta + 3 sin theta }/( 3 sin theta - sin 3theta )#
#- 1/2 ( 1 + cos 2theta ) -( 1 - cos 2theta )/( 1 + cos 2theta )#

Use

#cos 3A = 4 cos^3A - 3 cos A and sin 3A = 3 sin A - 4 sin^3A. #

#cot^3theta - cos^2theta - tan^2theta#

# = cos^3theta/sin^3theta - cos^2theta - sin^2theta/cos^2theta#

#= ( cos 3theta + 3 sin theta }/( 3 sin theta - sin 3theta )#

#- 1/2 ( 1 + cos 2theta ) -( 1 - cos 2theta )/( 1 + cos 2theta )#