sin(90-theta)sin theta/tan theta -1= -sin^2theta prove it
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sin(90-theta)=cos theta
cos *sin/(sin/cos)-1
(cos*sin*cos)/sin
cos square - 1
sin square+cos square=1
sin square=cos square - 1
=sin square theta
cos *sin/(sin/cos)-1
(cos*sin*cos)/sin
cos square - 1
sin square+cos square=1
sin square=cos square - 1
=sin square theta
Answered by
15
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