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[" 1."(1+sin theta-cos theta)/(1+sin the...

[" 1."(1+sin theta-cos theta)/(1+sin theta+cos theta)=],[[" 1) "sin(theta)/(2)quad " 2) "cos(theta)/(2)," 3) "tan(theta)/(2)]]

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If sin theta+cos theta=1 then sin theta cos theta=

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(sin theta-cos theta+1)/(sin theta+cos theta-1)=(1+sin theta)/(cos theta)

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

(1-cos theta)/(sin theta)=(sin theta)/(1+cos theta)

(1-cos theta)/(sin theta)=(sin theta)/(1+cos theta)

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