How do you express #tan theta - cot^2theta # in terms of #cos theta #?

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Answer 1 · 2016-01-16T06:21:13

Explanation is given below.

#tantheta - cot^2theta#

On handiling these kind of problem apply your previous knowledge on identity.

#tantheta = sintheta/costheta#

#cottheta = costheta/sintheta#

#sin^2theta = 1-cos^2theta#

Our problem:

#tantheta - cot^2theta#

#=sintheta/costheta - cos^2theta/sin^2theta#

#=sqrt(1-cos^2theta)/cos(theta) - cos^2theta/(1-cos^2theta)#

#=(sqrt(1-cos^2theta)(1-cos^2theta))/(costheta(1-cos^2theta)) -(cos^2thetacostheta)/(costheta(1-cos^2theta)#

#=((1-cos^2theta)^(3/2)-cos^3theta)/(costheta(1-cos^2theta))#