How do I prove the identity of (sin^2Theta+2cosTheta-1)/(sin^2Theta+3cosTheta-3)=(cos^2Theta+cosTheta)/(-sin^2Theta) ?

(sin^2Theta+2cosTheta-1)/(sin^2Theta+3cosTheta-3)=(cos^2Theta+cosTheta)/(-sin^2Theta)

2 Answers
May 1, 2018

Please see below.

Explanation:

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

(-cos^2theta+2costheta)/(-cos^2theta+3costheta-2)=(cos^2theta-2costheta)/(cos^2theta-3costheta+2)=

(costheta(costheta-2))/((costheta-2)(costheta-1))=costheta/(costheta-1)=(costheta(costheta+1))/((costheta-1)(costheta+1))

(cos^2theta+costheta)/(cos^2theta-1)=(cos^2theta+costheta)/-sin^2theta

May 1, 2018

Turn everything to cosines and grind out the algebra.

Explanation:

{ sin ^2 theta + 2 cos theta - 1}/ { sin ^2 theta + 3 cos theta - 3}

= { 1 - cos ^2 theta + 2 cos theta - 1}/ { 1 - cos ^2 theta + 3 cos theta - 3}

= { -cos theta (cos theta -2) }/ { -(cos ^2 theta - 3 cos theta + 2)}

= { cos theta (cos theta -2) }/ { ( cos theta - 2)(cos theta - 1)}

= { cos theta }/ { cos theta - 1 }

= { cos theta }/ { cos theta - 1 } cdot {cos theta + 1}/{cos theta + 1}

= { cos ^2theta + cos theta }/{cos^2 theta -1 }

= { cos ^2theta + cos theta }/{- sin ^ 2 theta } quad sqrt