Question #ded10

Question

1328 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-21T04:01:04

#(sin(pi + theta )+ cos( pi/2+theta) + cot( pi- theta))/tan((3pi)/2-theta)#

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

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

#=-2sin( theta)/cottheta -cot( theta)/cot(theta)#

#=-2sin( theta)/(costheta/sintheta) -1#

#=-2sin^2( theta)/costheta-1#

#=-(2(1-cos^2 theta))/costheta-1#

#=-2/costheta+(2cos^2theta)/costheta-1#

#=-2sectheta+2costheta-1#

#=2costheta-2sectheta-1#