Question

secθ+tanθ=3 then secθ=?​

Answer 1

Answer:

$tanx = 3 - secx \\ squaring \: both \: sides \\ {tan}^{2} x = 9 + {sec}^{2} x - 6secx \\ {sec}^{2} x - 1 = 9 + {sec}^{2} x - 6secx \\ 6secx = 10 \\ secx = \frac{10}{6} = \frac{5}{3}$