Answer
$sec(arctan~3x)=±\sqrt {1+9x^2}$
Work Step by Step
Let $u=arctan~3x~~$ (Range: $-\frac{\pi}{2}\lt u\lt\frac{\pi}{2}$)
Then: $3x=tan~u$
$sec^2u=1+tan^2u$
$sec^2u=1+(3x)^2=1+9x^2$
$sec~u=±\sqrt {1+9x^2}$
$sec(arctan~3x)=sec~u=±\sqrt {1+9x^2}$
