How do you simplify #cot x xx sin x#?

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Answer 1 · 2016-08-17T12:27:04

# (cos x)/(sin x) xx sinx#

=# cosx#

All of the 6 trig ratios can be expressed in terms of #sin theta # and / or #cos theta#

#tan theta = (sin theta)/(cos theta) " "(o/h)/(a/h) = (o xxh)/(axxh) = o/a#

#cot theta = 1/(tan theta) = (cos theta)/(sin theta)#

#sec theta = 1/(cos theta)#

#cosec theta =1/(sin theta)#

#cotx xxsinx #

=# (cos x)/cancel(sin x) xxcancel sinx#

=# cosx#