What is #sin^2theta-tantheta # in terms of #costheta#?

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Answer 1 · 2016-01-15T04:14:43

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

Write in terms of #sintheta# and #costheta#.

#=sin^2theta-sintheta/costheta#

Find a common denominator.

#=(sin^2thetacostheta)/costheta-sintheta/costheta#

Combine.

#=(sin^2thetacostheta-sintheta)/costheta#

Use the following identities:

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