6. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& \sin ^{2} x+\cos ^{2} y=\tan ^{2} z, \\
& \sin ^{2} y+\cos ^{2} z=\tan ^{2} x, \\
& \sin ^{2} z+\cos ^{2} x=\tan ^{2} y .
\end{aligned}
$$
SOLUTION. By substituting $\cos ^{2} x=a, \cos ^{2} y=b, \cos ^{2} z=c$, the system becomes

\[
\begin{aligned}
& 1-a+b=\frac{1}{c}-1, \\
& 1-b+c=\frac{1}{a}-1, \\
& 1-c+a=\frac{1}{b}-1,
\end{aligned}
\]

where $a, b, c \in(0,1\rangle$.

By adding these equations, we get

\[
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6
\]

so the harmonic mean of the numbers $a, b, c$ is $\frac{1}{2}$.

By multiplying the equations by $c, a, b$ respectively, we have

\[
\begin{aligned}
c-a c+b c & =1-c \\
a-a b+a c & =1-a \\
b-b c+a b & =1-b
\end{aligned}
\]

and by adding them, $2(a+b+c)=3$. Thus, the arithmetic mean of the numbers $a, b, c$ is also $\frac{1}{2}$. From the equality of the arithmetic and harmonic means, it follows that $a=b=c=\frac{1}{2}$. It is easy to verify that this triplet satisfies the system (1). The solutions to the originally given system are therefore all triplets $\left(\frac{1}{4} \pi+\frac{1}{2} k \pi, \frac{1}{4} \pi+\frac{1}{2} l \pi, \frac{1}{4} \pi+\frac{1}{2} m \pi\right)$, where $k, l, m$ are integers.