Recall that
tan²(<em>x</em>) + 1 = sec²(<em>x</em>)
for all <em>x</em> ≠ (2<em>n</em> + 1) 90°, where <em>n</em> is any integer. (That is, whenever <em>x</em> isn't an angle that is an odd multiple of 90°.)
Then
tan²(Φ) + sec²(Φ) = 3
tan²(Φ) + (tan²(Φ) + 1) = 3
2 tan²(Φ) + 1 = 3
2 tan²(Φ) = 2
tan²(Φ) = 1
tan(Φ) = ±√1
tan(Φ) = 1 <u>or</u> tan(Φ) = -1
Φ = tan⁻¹(1) + <em>n</em> 180° <u>or</u> Φ = tan⁻¹(-1) + <em>n</em> 180°
Φ = 45° + <em>n</em> 180° <u>or</u> Φ = -45° + <em>n</em> 180°
The first family of solution is the set of angles
{…, -315°, -135°, 45°, 225°, 405°, …}
and the second family is the set
{…, -405°, -225°, -45°, 135°, 315°, …}
(showing the solutions for <em>n</em> = -2 to <em>n</em> = 2)
You can condense the solution set into one family by noticing each angle is 45° plus some multiple of 90°, so that
Φ = 45° + <em>n</em> 90°
If you're looking for solutions in a given range, such as 0 ≤ Φ < 360°, then
Φ = 45°, 135°, 225° or 315°
