# Sec A + Tan A = 3 => Sec A =1) 10/3 2) 5/3 3) 2/3 4) 4/3

manna
59 Points
5 years ago
$\fn_cm secA+tanA=3$
$\fn_cm 1/cosA+sinA/cosA=3$
1+sinA/cosA=3
1+sinA=3*cosA
1+sinA=3(I'll rewrite  as

Squaring both sides,

Let u=sinA. It is easy to see that we can now obtain a quadratic:

-->
= -1 or 4/5.
sinA=4/5
cosA=3/5
sec=1/cos=5/3