Question

Find sec(x) and csc(x) if tan(x) = 3 and cos > 0

Answer 1

Recall that

$1+\tan^2x=\sec^2x\implies\sec x=\sqrt{1+\tan^2x}$

We know to take the positive root here because $\cos x>0$, which means $\sec x>0$ as well. Then

$\sec x=\sqrt{1+3^2}=\sqrt4=2\implies\cos x=\dfrac12$

We then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\sin x}{\frac12}=3\implies\sin x=\dfrac32\implies\sec x=\dfrac23$