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

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Answer 1 · 2018-02-24T06:34:07

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

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

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

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

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

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

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

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

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

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

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