Question

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

Answer 1

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))`