Fall Mathematical Competition

Let $x=\tan \alpha$, $y=\tan \beta$.

We have ${\cos }^2\alpha =\frac{1}{1+x^2}$ and ${\cos }^2\beta =\frac{1}{1+y^2}$.

So, $$\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)(1+xy)=2.$$

Multiply both sides by $(1+x^2)(1+y^2)$: $$\left[(1+y^2)+(1+x^2)\right](1+xy)=2(1+x^2)(1+y^2)$$ $$(2+x^2+y^2)(1+xy)=2(1+x^2+y^2+x^2{y}^2)$$

Expand the left side: $$(2+x^2+y^2)(1+xy)=2(1+xy)+x^2(1+xy)+y^2(1+xy)$$ $$=2(1+xy)+x^2+x^{3}y+y^2+x{y}^3$$

But let's expand both sides fully:

Left: $$(2+x^2+y^2)(1+xy)=2(1+xy)+x^2(1+xy)+y^2(1+xy)$$ $$=2+2xy+x^2+x^{3}y+y^2+x{y}^3$$

Right: $$2(1+x^2)(1+y^2)=2(1+x^2+y^2+x^2{y}^2)=2+2x^2+2y^2+2x^2{y}^2$$

Set equal: $$2+2xy+x^2+x^{3}y+y^2+x{y}^3=2+2x^2+2y^2+2x^2{y}^2$$

Subtract $2$ from both sides: $$2xy+x^2+x^{3}y+y^2+x{y}^3=2x^2+2y^2+2x^2{y}^2$$

Bring all terms to one side: $$2xy+x^2+x^{3}y+y^2+x{y}^3-2x^2-2y^2-2x^2{y}^2=0$$ $$2xy+x^2+x^{3}y+y^2+x{y}^3-2x^2-2y^2-2x^2{y}^2=0$$ Group terms: $$(x^2-2x^2)+(y^2-2y^2)+2xy+x^{3}y+x{y}^3-2x^2{y}^2=0$$ $$(-x^2)+(-y^2)+2xy+x^{3}y+x{y}^3-2x^2{y}^2=0$$

Now, $x^{3}y+x{y}^3=xy(x^2+y^2)$, so: $$(-x^2)+(-y^2)+2xy+xy(x^2+y^2)-2x^2{y}^2=0$$ $$(-x^2-y^2)+2xy+xy(x^2+y^2)-2x^2{y}^2=0$$

Now, $xy(x^2+y^2)=x^{3}y+x{y}^3$ as above.

Group $-x^2-y^2$ and $xy(x^2+y^2)$: $$2xy+xy(x^2+y^2)-x^2-y^2-2x^2{y}^2=0$$

Let us factor $xy$: $$2xy+xy(x^2+y^2)=xy[2+x^2+y^2]$$ So: $$xy[2+x^2+y^2]-x^2-y^2-2x^2{y}^2=0$$

Bring $x^2+y^2$ to the other side: $$xy[2+x^2+y^2]=x^2+y^2+2x^2{y}^2$$

Now, move all terms to one side: $$xy[2+x^2+y^2]-x^2-y^2-2x^2{y}^2=0$$

Alternatively, let's try the hint: write the given equality as $(\tan \alpha \tan \beta -1)(\tan \alpha -\tan \beta {)}^2=0$.

Let us try to factor the expression.

Let $x=\tan \alpha$, $y=\tan \beta$.

Suppose $xy-1=0$, i.e. $xy=1$.

Then $\tan \alpha \tan \beta =1$.

Recall that $\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$.

If $\tan \alpha \tan \beta =1$, then denominator is $0$, so $\tan (\alpha +\beta )$ is undefined, which happens when $\alpha +\beta =90^{\circ }$.

Since $\alpha$ and $\beta$ are acute and $\alpha \ne \beta$, this is possible.

Alternatively, $(x-y{)}^2=0$ gives $x=y$, i.e. $\tan \alpha =\tan \beta$, so $\alpha =\beta$, but $\alpha \ne \beta$ by assumption.

Therefore, the only possibility is $\tan \alpha \tan \beta =1$, i.e. $\alpha +\beta =90^{\circ }$.

Thus, $\boxed{\alpha +\beta =90^{\circ }}$.