How do I simplify #sinThetacotTheta-sinThetacosTheta# ?

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Answer 1 · 2018-05-01T02:18:08

#sin theta cot theta - sin theta cos theta ## = cos theta (1 - sin theta)## = cos theta - 1/2 sin(2 theta) #

#sin theta cot theta - sin theta cos theta #

# = sin theta (cos theta / sin theta) - sin theta cos theta #

# = cos theta - sin theta cos theta #

# = cos theta (1 - sin theta)#

There are other ways we could have gone, say

# = cos theta - sin theta cos theta #

# = cos theta - 1/2 sin(2 theta) #