How do you express #sin theta - cottheta + tan theta # in terms of #cos theta #?

1 Answer
May 29, 2016

#=sintheta - costheta/sintheta + sin theta/costheta#

#=(sin theta xx sin theta xx costheta)/(sin theta xx costheta) + (costheta xx costheta)/(costhetasintheta) + (sintheta xx sin theta)/(costhetasintheta)#

#=(sin^2thetacostheta + cos^2theta + sin^2theta)/(sinthetacostheta)#

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

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

#=sqrt(((1 - cos^2theta)/sintheta)^2)#

#=sqrt((1 - 2cos^2theta + cos^4theta)/(sin^2theta))#

#=sqrt((1 - 2cos^2theta + cos^4theta)/(1 - cos^2theta))#

#=sqrt((-(cos2theta) + cos^4theta)/(1 - cos^2theta)#

Certainly a long proof, but it works, thankfully!

Hopefully this helps, and have a great day!