Source: 2007 Bulgarian Autumn Math Competition, Problem 11.1
Tags: algebra, Trigonometric Equations
Let $0<\alpha,\beta<\frac{\pi}{2}$ which satisfy
\[(\cos^2\alpha+\cos^2\beta)(1+\tan\alpha\tan\beta)=2\]Prove that $\alpha+\beta=\frac{\pi}{2}$.
$(\cos^{2}\alpha+\cos^{2}\beta)(1+\tan\alpha\tan\beta)=2$.
$(\frac{1+\cos 2\alpha}{2}+\frac{1+\cos 2\beta}{2})(\cos \alpha\cos \beta+\sin \alpha\sin \beta)=2\cos \alpha\cos \beta$.
$(\frac{2+\cos 2\alpha+\cos 2\beta}{2})\cos(\alpha-\beta)=\cos(\alpha+\beta)+\cos(\alpha-\beta)$.
$(\frac{2+2\cos(\alpha+\beta)\cos(\alpha-\beta)}{2})\cos(\alpha-\beta)=\cos(\alpha+\beta)+\cos(\alpha-\beta)$.
$(1+\cos(\alpha+\beta)\cos(\alpha-\beta)\ )\cos(\alpha-\beta)=\cos(\alpha+\beta)+\cos(\alpha-\beta)$.
$\cos(\alpha-\beta)+\cos(\alpha+\beta)\cos^{2}(\alpha-\beta)=\cos(\alpha+\beta)+\cos(\alpha-\beta)$.
$\cos(\alpha+\beta)\cos^{2}(\alpha-\beta)=\cos(\alpha+\beta)$.
$0=\cos(\alpha+\beta)\sin^{2}(\alpha-\beta)$.