Prove $\tan(a+b)$ formula by using $\arctan(x)$ Hey guys have an exam tomorrow and I have been practicing a no. of questions yet I cannot figure this one out.
Knowing
$$\arctan(x) = \int_{0}^{x} \frac{1}{1 + w^2}\text{d}w$$
prove that
$$\tan(x+y) = \frac{\tan x + \tan y}{ 1 - \tan x \tan y}$$
between $-\pi/2$ and $\pi/2$
I have tried using arctan(x) + arctan(1/x) = pi/2 as a way of expressing how the a and b could impact one another. After that though I'm stumped. Maybe doing nothing for the past 12 hours except studying isn't the greatest idea.
 A: By subbing $x=\arctan u$  and  $y=\arctan t$  into the desired identity, we see that proving the tangent identity is equivalent to proving that 
$$\arctan u+\arctan t=\arctan\left(\frac{u+t}{1-ut}\right)+(0,~ \pi,~ or~-\pi)$$
This is true if both sides of this are equal for one particular $u$ in each interval where both sides are continuous and if, for any $t$, the derivative of both sides with respect to $u$ is equal for all $u$. This first condition can be verified by taking the limit $u\to 1/t$; the LHS is $\pm \pi/2$, as is the RHS (though the sign will be different for the right and left-handed limits). This also deals iwth the jump discontinuity in the RHS.
To check the derivative condition, we can use the given integral identity to find that
$$\frac{d}{du}(\arctan u+\arctan t)=\frac{1}{1+u^2}$$
and 
$$\frac{d}{du}\arctan\left(\frac{u+t}{1-ut}\right)=\frac{1
+t^2}{(1-ut)^2+(u+t)^2}$$
Now, using plain old algebra,
$$\frac{1}{1+u^2}=\frac{1
+t^2}{(1-ut)^2+(u+t)^2}$$
$$\Longleftrightarrow$$
$$(1+u^2)(1+t^2)=(1-ut)^2+(u+t)^2$$
which can be verified by expanding both sides. Hence the derivatives of both sides are equal and by the earlier statements, we are done.
