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sin^2 theta + 2 cos theta -1/ (sin^2 theta +3 cos theta-3= cos^2 theta+cos theta/ (-sin^2 theta) Prove the identity

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

Prove the identity.
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Start from the left, it would be easy to change sin²(θ) into cos²(θ) using the identity:
sin²(θ)+cos²(θ)=1

Do not forget to put parentheses around the numerator and denominator. The expression that you posted does not evaluate to what you meant.

(sin²(θ)+2cos(θ)-1) / (sin²(θ)+3cos(θ)-3)
=(2cos(θ)-cos²(θ)) /
(3cos(θ)-cos²(θ)-2)
=cos(θ)(2-cos(θ)) / ((cos(θ)-2)(1-cos(θ)))
=cos(θ)/(cos(θ)-1)

multiply by (1+cos(θ)) both numerator and denominator:

= cos(θ)(1+cos(θ)) / ((cos(θ)-1)(1+cos(θ)))
=-cos(θ)(1+cos(θ)) / sin²(θ)
answered by: latha
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