How do you simplify #cos(pi-x)+cotxsin(x-pi/2)#?

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Answer 1 · 2016-11-04T01:17:07

The expression can be simplified to #-cotx(sinx + 1 )#.

Use the following identities.

#cos(A - B) = cosAcosB + sinAsinB#
#sin(A - B) = sinAcosB - cosAsinB#

#=>cospicosx + sinpisinx + cotx(sinxcos(pi/2) - cosxsin(pi/2))#

#=>-cosx + 0 + cosx/sinx(0 - cosx)#

#=>-cosx - cos^2x/sinx#

#=>(-cosxsinx - cos^2x)/sinx#

#=>(-cosx(sinx + cosx))/sinx#

#=>-cotx(sinx +cosx)#

Hopefully this helps!