This proof that $\pi=0$ may be of some interest in examinations. 

The function $f(x)=\arctan(x)+\arctan(1/x)$ has derivative $f’(x)=\frac1{1+x^2}-\frac1{x^2} \frac{1}{1+\frac1{x^2}}=0$, hence it is constant. Therefore$\displaystyle \lim_{x\to+\infty}f(x)= \displaystyle \lim_{x\to-\infty}f(x)$, that is $\frac\pi2=-\frac\pi2$, whence $\pi=0$.$\quad\square$