Question

If x, y and z are real numbers that satisfy the three equations

$\begin{cases} \tan(x) + \tan(y) + \tan(z) = 6 - (\cot(x) + \cot(y) + \cot(z)) \\ \tan^2(x) + \tan^2(y) + \tan^2(z) = 6 - (\cot^2(x) + \cot^2(y) + \cot^2(z)) \\ \tan^3(x) + \tan^3(y) + \tan^3(z) = 6 - (\cot^3(x) + \cot^3(y) + \cot^3(z)) \end{cases}$

Find the value of the expression $(\frac{\tan(x)}{\tan(y)} + \frac{\tan(y)}{\tan(z)} + \frac{\tan(z)}{\tan(x)} + \frac{\tan(y)}{\tan(x)} + \frac{\tan(z)}{\tan(y)} + \frac{\tan(x)}{\tan(z)} + 3\tan(x)\tan(y)\tan(z))$.

Answer 1

9

Answer 2

Let $a = \tan(x)$, $b = \tan(y)$, and $c = \tan(z)$. Then $\cot(x) = \frac{1}{a}$, $\cot(y) = \frac{1}{b}$, and $\cot(z) = \frac{1}{c}$.

The given system of equations can be rewritten as:

1. $a + b + c = 6 - (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
2. $a^2 + b^2 + c^2 = 6 - (\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2})$
3. $a^3 + b^3 + c^3 = 6 - (\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3})$

Rearranging these equations by grouping terms:

1. $(a + \frac{1}{a}) + (b + \frac{1}{b}) + (c + \frac{1}{c}) = 6$
2. $(a^2 + \frac{1}{a^2}) + (b^2 + \frac{1}{b^2}) + (c^2 + \frac{1}{c^2}) = 6$
3. $(a^3 + \frac{1}{a^3}) + (b^3 + \frac{1}{b^3}) + (c^3 + \frac{1}{c^3}) = 6$

Let's introduce new variables: $u = a + \frac{1}{a}$ $v = b + \frac{1}{b}$ $w = c + \frac{1}{c}$

We know the following identities: $t^2 + \frac{1}{t^2} = (t + \frac{1}{t})^2 - 2$ $t^3 + \frac{1}{t^3} = (t + \frac{1}{t})^3 - 3(t + \frac{1}{t})$

Using these identities, the system of equations in terms of $u, v, w$ becomes:

1. $u + v + w = 6$
2. $(u^2 - 2) + (v^2 - 2) + (w^2 - 2) = 6$ $u^2 + v^2 + w^2 - 6 = 6$ $u^2 + v^2 + w^2 = 12$
3. $(u^3 - 3u) + (v^3 - 3v) + (w^3 - 3w) = 6$ $u^3 + v^3 + w^3 - 3(u + v + w) = 6$ Substitute $u+v+w=6$ from the first equation: $u^3 + v^3 + w^3 - 3(6) = 6$ $u^3 + v^3 + w^3 - 18 = 6$ $u^3 + v^3 + w^3 = 24$

Now we have a system of equations for $u, v, w$: I. $u + v + w = 6$ II. $u^2 + v^2 + w^2 = 12$ III. $u^3 + v^3 + w^3 = 24$

Let $e_1 = u+v+w$, $e_2 = uv+vw+wu$, and $e_3 = uvw$. From (I), $e_1 = 6$.

We know that $(u+v+w)^2 = u^2+v^2+w^2 + 2(uv+vw+wu)$. Substitute values from (I) and (II): $6^2 = 12 + 2e_2$ $36 = 12 + 2e_2$ $2e_2 = 24 \implies e_2 = 12$.

Now, use the identity for the sum of cubes: $u^3+v^3+w^3 - 3uvw = (u+v+w)(u^2+v^2+w^2 - (uv+vw+wu))$ Substitute values from (I), (II), (III), and the calculated $e_2$: $24 - 3e_3 = (6)(12 - 12)$ $24 - 3e_3 = 6 \cdot 0$ $24 - 3e_3 = 0$ $3e_3 = 24 \implies e_3 = 8$.

So, $u, v, w$ are the roots of the cubic polynomial equation $t^3 - e_1 t^2 + e_2 t - e_3 = 0$: $t^3 - 6t^2 + 12t - 8 = 0$ This equation can be recognized as the expansion of $(t-2)^3$: $(t-2)^3 = t^3 - 3(t^2)(2) + 3(t)(2^2) - 2^3 = t^3 - 6t^2 + 12t - 8$. Thus, $(t-2)^3 = 0$, which implies $t=2$ is the only solution.

Therefore, $u=2, v=2, w=2$.

Now we substitute back to find $a, b, c$: $u = a + \frac{1}{a} = 2$ $a^2 + 1 = 2a$ $a^2 - 2a + 1 = 0$ $(a-1)^2 = 0 \implies a=1$.

Similarly, $b + \frac{1}{b} = 2 \implies b=1$. And $c + \frac{1}{c} = 2 \implies c=1$.

So, $\tan(x) = 1$, $\tan(y) = 1$, $\tan(z) = 1$.

Finally, we need to find the value of the expression: $(\frac{\tan(x)}{\tan(y)} + \frac{\tan(y)}{\tan(z)} + \frac{\tan(z)}{\tan(x)} + \frac{\tan(y)}{\tan(x)} + \frac{\tan(z)}{\tan(y)} + \frac{\tan(x)}{\tan(z)} + 3\tan(x)\tan(y)\tan(z))$

Substitute $a=1, b=1, c=1$ into the expression: $(\frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + 3(1)(1)(1))$ $= (1 + 1 + 1 + 1 + 1 + 1 + 3)$ $= 6 + 3$ $= 9$.

The final answer is $\boxed{9}$.

Explanation of the solution:

1. Transform the given equations using $a=\tan x, b=\tan y, c=\tan z$ and their reciprocals.
2. Define new variables $u = a + \frac{1}{a}$, $v = b + \frac{1}{b}$, $w = c + \frac{1}{c}$.
3. Rewrite the system in terms of $u, v, w$ using identities for $t^2+\frac{1}{t^2}$ and $t^3+\frac{1}{t^3}$. This yields: $u+v+w=6$ $u^2+v^2+w^2=12$ $u^3+v^3+w^3=24$
4. Recognize that $u,v,w$ are roots of a cubic polynomial $t^3 - e_1 t^2 + e_2 t - e_3 = 0$, where $e_1, e_2, e_3$ are elementary symmetric polynomials.
5. Calculate $e_1=6$, $e_2=12$, $e_3=8$ using the sums of powers.
6. Form the cubic equation $t^3 - 6t^2 + 12t - 8 = 0$, which simplifies to $(t-2)^3 = 0$.
7. This implies $u=v=w=2$.
8. Solve $a + \frac{1}{a} = 2$ to find $a=1$. Similarly, $b=1$ and $c=1$.
9. Substitute $a=1, b=1, c=1$ into the target expression to get the final numerical value.