Can I get some help please?
1 Answer
Sep 26, 2017
We have
#sintheta/costheta + 1/tantheta = 1/(sinthetacostheta)#
Using
#sintheta/costheta + 1/(sintheta/costheta) = 1/(sinthetacostheta)#
#sintheta/costheta + costheta/sintheta = 1/(sinthetacostheta)#
#(sintheta(sin theta) + costheta(costheta))/(costhetasintheta) = 1/(costhetasintheta)#
#(sin^2theta + cos^2theta)/(costhetasintheta)= 1/(costhetasintheta)#
Because
#1/(costhetasintheta) = 1/(costhetasintheta)#
Hopefully this helps!