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

Question

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

1706 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-01T04:14:17

Please see below.

.

#(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#

Answer 2 · 2018-05-01T04:14:25

Turn everything to cosines and grind out the algebra.

# { 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#