Recall the definitions:

- #sec(x)=1/\cos(x)#
- #\tan(x)=\sin(x)/\cos(x)#

We can thus write #sec^2(x) + 2 \tan^2(x)# as

#1/\cos^2(x) + 2 \sin^2(x)/\cos^2(x)= {1+2\sin^2(x)}/{\cos^2(x)}#.

We want this expression to equal #4#, so we can multiply for #\cos^2(x)# and get

#{1+2\sin^2(x)}/{\cos^2(x)}=4 \iff 1+2\sin^2(x)=4\cos^2(x)#.

Subtracting #\cos^2(x)# from both sides, we have

#1-\cos^2(x) +2\sin^2(x)=3\cos^2(x)#,

and since #1-\cos^2(x)# equals #\sin^2(x)#, the expression becomes

#3\sin^2(x)=3\cos^2(x)#.

Dividing the whole expression by the right member, we have

#{3\sin^2(x)}/{3\cos^2(x)}=1#. Canceling the #3#'s out, and recalling again that #\tan(x)=\sin(x)/\cos(x)#, we finally find out that the previous equation is the same as

#\tan^2(x)=1#, which is verified (considering only the angles between #0# and #2\pi#) when #\tan(x)=1#, i.e. when #x=\pi/4#, and when #\tan(x)=-1#, i.e. when #x={5\pi}/4#