Answer
$$\tan x=\pm\sqrt{\sec^2 x-1}$$
Work Step by Step
Pythagorean Identities:
$$\tan^2 x+1=\sec^2 x$$
Therefore,
$$\tan^2 x=\sec^2 x-1$$
We now take the square root of both sides to get $\tan x$
$$\sqrt{\tan^2 x}=\sqrt{\sec^2 x-1}$$
$$\tan x=\pm\sqrt{\sec^2 x-1}$$
(Do not forget the $\pm$ sign, since $\tan x$ might be positive or negative)
