How do you simplify #sin(theta+pi)/cos(theta-pi)#?
1 Answer
Dec 7, 2016
This simplifies to
Explanation:
Use the formulas
#=>(sinthetacospi + costhetasinpi)/(costhetacospi + sinthetasinpi)#
#=>(sintheta(-1) + costheta(0))/(costheta(-1) + sin theta(0))#
#=> (-sintheta)/(-costheta)#
Use the identity that
#=> tantheta#
Hopefully this helps!