## Trigonometry (11th Edition) Clone

$$\tan x=\pm\sqrt{\sec^2 x-1}$$
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)