How do you prove #sin(2theta)/sintheta-cos(2theta)/costheta=sectheta#?

Question

9911 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-05T19:32:27

see below

#sin(2theta)/sin theta - cos(2theta)/cos theta = sec theta#

Left Side: #=sin(2theta)/sin theta - cos(2theta)/cos theta#

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

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

#=2cos theta-2cos theta+1/cos theta#

#=1/cos theta#

#=sec theta#

#=# Right Side