In answer #1, if \(\cos x = 0,\) then \(\sec x\) will be infinite and so should be rejected.
Here are two alternative routines.
\(\tan x + \sec x = 2, \\ \sec x =2 - \tan x, \\ \sec^{2}x=(2 - \tan x)^{2}, \\ 1+ \tan^{2}x = 4-4\tan x+\tan^{2}x, \\ 4 \tan x = 3, \\ \tan x = 3/4. \)
The 3 and the 4 can be seen as two legs of a 3, 4, 5 rt-angled triangle from which we can read off the other ratios.
\(\sin x = 3/5, \cos x = 4/5,\sec x = 5/4,\)
and there are two others.
Note that \(\tan x + \sec x = 3/4+5/4 = 8/4 =2.\)
Alternatively,
\(\tan x + \sec x = \frac {\sin x}{ \cos x }+\frac{1}{\cos x} = 2,\\ \sin x + 1 = 2\cos x, \\ (\sin x+1)^{2}=4\cos^{2}x,\\ \sin^{2}x+2\sin x + 1 = 4(1-\sin^{2} x ),\\ 5\sin^{2}x+2\sin x -3=0,\\ (5\sin x-3)(\sin x +1)=0, \)
\(5\sin^{2}x +2\sin x -3 = 0, \\(5\sin x -3)(\sin x +1)=0,\\ \sin x = 3/5,\)
etc.
(the second root sin x = -1 is rejected because that implies that cos x = 0 meaning sec x is infinite.)
