Show that $\arctan a+\arctan(\frac{1}{a})=\frac{\pi}{2}$ $$\arctan a+\arctan\left(\frac 1 a \right)=\frac \pi 2$$
I have the mark scheme in front of me, and I understand where the numbers come from, but I don't understand why they do what they do.
You need this part in to show it:

The markscheme says:

Again, I understand where the numbers come from, but can someone explain to me where the first line comes from, as well as the rest of it? It's a "Hence" question so I have to solve it this way.
 A: Notice that the derivative of $\arctan x$ is $\frac1{1+x^2}$ so we have
$$\int_1^\alpha \frac{dx}{1+x^2}=\arctan x\Bigg|_1^\alpha$$
now use the change of variable $t=\frac1x$ to find the given equality. Can you take it from here?
A: This is too long for a comment, but here's a method that uses only geometry and the definition of $\arctan$: For $a > 0$, draw a right triangle with legs $1$ and $a$ adjacent to the right angle. Then, the two acute angles are, by definition $\arctan a$ and $\arctan \left( \frac{1}{a} \right)$, and since the triangle is right, the sum of these two angles is $\frac{\pi}{2}$.
Remark Note that your identity holds only for positive $a$. For negative $a$ the sum is equal to $-\frac{\pi}{2}$, which you can show by applying the identity to $|a|$ and using that $\arctan$ is odd.
A: Differentiate with respect to $a$, to prove that it has to be constant. Then taking the value at $a=1$ gives you the result. That's probably the easiest way to prove it.
A: Supose $a\neq 0$. It's just a way to remember:$$\tan\left(\arctan a+\arctan\frac{1}{a}\right)=\frac{\tan(\arctan a)+\tan\left(\arctan\frac{1}{a}\right)}{1-\tan(\arctan a)\arctan\left(\frac{1}{a}\right)}=\frac{a+\frac{1}{a}}{1-a\cdot \frac{1}{a}}=+\infty. $$ Moreover, $\arctan(+\infty )=\frac{\pi}{2}$
This proof is not correct (but funny anyway :-) )
