Use $T(x,y)=\begin{pmatrix}\frac{\sin x}{\cos y}\\ \frac{\sin y}{\cos x}\end{pmatrix}$ to evaluate $\sum_{n=1}^{\infty}\frac{1}{n^2}$ Let $O=\{(x,y) \in \mathbb R^2: x,y>0, x+y<\frac{\pi}{2}\}$ and $P=(0,1)\times (0,1)$.
Then $T:O \to \mathbb R^2, T(x,y)=\begin{pmatrix}\frac{\sin x}{\cos y}\\ \frac{\sin y}{\cos x}\end{pmatrix}$ is a diffeomorphism with $T(O)=P$
How can I use $T$ to evaluate $\sum_{n=1}^{\infty}\frac{1}{n^2}$?
The hints of the exercise are:
$3\sum_{n=1}^{\infty}\frac{1}{n^2}=4\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$ and for $n \in \mathbb N \cup \{0\}$: $\frac{1}{(n+1)^2}=\int_{(0,1)\times(0,1)}(xy)^n d\lambda_2(x,y)$

Well, 
$$ \frac{1}{(n+1)^2}=\iint_{(0,1)^2}(xy)^n\,dx\,dy $$
implies
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^2}=\iint_{(0,1)^2}\frac{dx\,dy}{1-x^2 y^2}.$$
Since the diffeomorphism $T$ maps the triangle $O=\{(a,b):a,b>0,a+b<\frac{\pi}{2}\}$ into the square $P=(0,1)^2$, by letting $x=\frac{\sin a}{\cos b},y=\frac{\sin b}{\cos a}$ the RHS of the previous line is converted (considering the Jacobian) into
$$ \iint_{O} \frac{1-\tan^2(a)\tan^2(b)}{1-\tan^2(a)\tan^2(b)}\,da\,db $$
so $\sum_{n\geq 0}\frac{1}{(2n+1)^2}$ turns out to be just the area of $O$, i.e. $\frac{\pi^2}{8}$. After this we get
$$ \zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}=\sum_{m\geq 1}\frac{1}{(2m)^2}+\sum_{m\geq 0}\frac{1}{(2m+1)^2} = \frac{1}{4}\zeta(2)+\frac{\pi^2}{8}$$
so we have a solution of Basel problem, $\zeta(2)=\color{red}{\frac{\pi^2}{6}}$.
