How do you express sin theta - cot theta + tan^2 theta sinθ−cotθ+tan2θ in terms of cos theta cosθ?

1 Answer
Jun 7, 2016

(1-cos theta-cos^2 theta)/sqrt(1-cos^2 theta)+1/cos^2 theta-11−cosθ−cos2θ√1−cos2θ+1cos2θ−1

Explanation:

Only Pythagotean identity is used here:
sin^2 theta+cos^2 theta=1sin2θ+cos2θ=1

sin theta-cot theta +tan^2 theta=sin theta-cos theta/sin theta+sin^2 theta/cos^2 theta=(sin^2 theta-cos theta)/sin theta+(1-cos^2 theta)/cos^2 theta=(1-cos theta-cos^2 theta)/sqrt(1-cos^2 theta)+1/cos^2 theta-1sinθ−cotθ+tan2θ=sinθ−cosθsinθ+sin2θcos2θ=sin2θ−cosθsinθ+1−cos2θcos2θ=1−cosθ−cos2θ√1−cos2θ+1cos2θ−1

I think it's the simplest form.

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