Question

How do you prove #(2-sec^2[theta])/(sec^2[theta]) =1-2sin^2[theta]#?

Answer 1

see below.

Explanation:

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

`(2-sec^2 theta )/(sec^2 theta ) = (2 cos^2theta-1)/cos^2theta cos^2theta = 2cos^2theta-1=2(1-sin^2theta)-1 = 1-2sin^2theta`