Originally Posted by

**Jhevon** I'd consider (b) as separate problem. Here is the trick. Let $\displaystyle \theta = \tan^{-1}x \Longrightarrow x = \frac x1 = \tan \theta$

So now you can proceed algebraically using trig identities, or geometrically and make your own right triangle. Tangent = opposite over adjacent, and so you can make a right triangle where one of the acute angles is $\displaystyle \theta$, the opposite side is $\displaystyle x$ and the adjacent side is $\displaystyle 1$. You can use Pythagoras to find the other side. With the fully populated triangle, you can find all trig functions of $\displaystyle \theta$. Would you know how to proceed?