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(cos^2x+1+sinx)/(cos^2x+3)=(1+sinx)/(2+sinx)

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2 Answers

If you want to prove from the left to the right, here is one way:
(cos^2x+1+sinx)/(cos^2x+3)
= (1-sin^2x + 1+sinx)/(1-sin^2x+3)
= (1+sinx)(1-sinx+1)/[(2+sinx)(2-sinx)]
= (1+sinx)/(2+sinx), after canceling 2-sinx.
(1+sin(x))/(2+sin(x))=(1+sin(x))*(2-sin(x))/[(2+sin(x))*(2-sin(x))]=(2+sin(x)-sin2(x))/(4-sin2(x))=
=[2+sin(x)-(1-cos2(x))]/[4-(1-cos2(x))]=(cos2(x)+1+sin(x))/(cos2(x)+3)