If sin theta = tan theta by2 find sin 3 theta
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solving LHS
=2 sin theta -2 sin ^3 theta/2cos ^3theta - cos theta
=sin theta(1-2 sin^2 theta) /cos theta (2 cos ^2 theta -1)
=tan theta ×[1-2(1-cos^2 theta) ]
[therefore sin^2theta=1-cos ^2 theta)
=tan theta ×(2cos^2 theta -1)/ ( 2 cos^ 2 theta-1)
=tan theta
therefore LHS = RHS
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