What is #csc^2theta-sec^2theta+cos^3theta# in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-09-24T08:36:55

Using the following identities here

1) #sin^2theta=1/2(1-cos2theta)#

2) #cos^2theta=1/2(1+cos2theta)#

3) #cos^3theta=1/4(cos3theta+3costheta)#

#csc^2theta-sec^2theta+cos^3theta#

#=1/sin^2theta-1/cos^2theta+cos^3theta#

#=2/(1-cos2theta)-2/(1+cos2theta)+1/4(cos3theta+3costheta)#