Simplify? #1/2 (tantheta-cottheta)/(tantheta+cottheta+1)#

1 Answer

I ended up with #-cos(2x)/(2+sin(2x))#

Explanation:

#1/2 (tantheta-cottheta)/(tantheta+cottheta+1)#

#(sintheta/costheta-costheta/sintheta)/(2(sintheta/costheta+costheta/sintheta+1)#

#((sin^2theta-cos^2theta)/(sinthetacostheta))/((2sin^2theta+2cos^2theta+2sinthetacostheta)/(sinthetacostheta))#

#((sin^2theta-cos^2theta)/(sinthetacostheta))xx((sinthetacostheta)/(2sin^2theta+2cos^2theta+2sinthetacostheta))#

#(sin^2theta-cos^2theta)/(2sin^2theta+2cos^2theta+2sinthetacostheta)#

Recall that #sin^2x+cos^2x=1

#(sin^2theta-cos^2theta)/(2+2sinthetacostheta)#

Recall that #cos(2x)=cos^2x-sin^2x=>-cos(2x)=sin^2x-cos^2x#

#-cos(2x)/(2+sin(2x))#