Question #f83fd

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Answer 1 · 2016-10-15T16:08:08

see below

$sin 2 theta=2 cot theta - (2cos^2 theta)/tan theta$

RHS: $2 cot theta - (2cos^2 theta)/tan theta=2/tan theta - 2cos^2 theta/ tan theta$

$=(2(1-cos^2 theta))/tan theta$

$=(2sin^2 theta)/tan theta$

$=(2 sin^2 theta) / (sin theta/ cos theta)$

$=2 sin^2 theta * cos theta/ sin theta$

$=2 sin theta cos theta$

$=sin 2 theta$

$=$ LHS