Sometimes, its better to adopt a complex way than a simpler one

I will elaborate the solution very clearly.

Since $\color{#D61F06}{\cos(x) + \cos(y) + \cos(z)} = \color{#20A900}{\sin(x) + \sin(y) + \sin(z)} = 0$ .

$\implies \left(\color{#D61F06}{\cos(x) + \cos(y) + \cos(z)} \right) + i \left(\color{#20A900}{\sin(x) + \sin(y) + \sin(z)} \right) = 0 \\ \implies \left(\color{#856d4d}{\cos(x) + i \sin(x)}\right) + \left(\color{#3D99F6}{\cos(y) + i\sin(y)} \right)+ \left(\color{#BA33D6}{\cos(z)+ i\sin(z)} \right) = 0 \\ \implies \left(\color{#856d4d}{e^{ix}}\right) + \left(\color{#3D99F6}{e^{iy}} \right)+ \left(\color{#BA33D6}{e^{iz}} \right) = 0$

Similarly,

$\implies \left(\color{#D61F06}{\cos(x) + \cos(y) + \cos(z)} \right) - i \left(\color{#20A900}{\sin(x) + \sin(y) + \sin(z)} \right) = 0 \\ \implies \left(\color{#856d4d}{\cos(x) - i \sin(x)}\right) + \left(\color{#3D99F6}{\cos(y) - i\sin(y)} \right)+ \left(\color{#BA33D6}{\cos(z)- i\sin(z)} \right) = 0 \\ \implies \left(\color{#856d4d}{e^{-ix}}\right) + \left(\color{#3D99F6}{e^{-iy}} \right)+ \left(\color{#BA33D6}{e^{-iz}} \right) = 0$

Let $\color{#856d4d}{a=e^{ix}}$ , $\color{#3D99F6}{b=e^{iy}}$ and $\color{#BA33D6}{c=e^{iz}}$ .

So, the obtained equations transform into $\color{#856d4d}{a} + \color{#3D99F6}{b} + \color{#BA33D6}{c} = 0$ and $\color{#856d4d}{\dfrac{1}{a}} + \color{#3D99F6}{\dfrac{1}{b}} + \color{#BA33D6}{\dfrac{1}{c}} = 0$ .

Consider the second equation,

$\color{#856d4d}{\dfrac{1}{a}} + \color{#3D99F6}{\dfrac{1}{b}} + \color{#BA33D6}{\dfrac{1}{c}} = 0 \\ \implies \dfrac{\color{#856d4d}{a} \color{#3D99F6}{b} + \color{#3D99F6}{b} \color{#BA33D6}{c} + \color{#BA33D6}{c} \color{#856d4d}{a}}{\color{#856d4d}{a} \color{#3D99F6}{b} \color{#BA33D6}{c}} = 0 \\ \implies \color{#856d4d}{a} \color{#3D99F6}{b} + \color{#3D99F6}{b} \color{#BA33D6}{c} + \color{#BA33D6}{c} \color{#856d4d}{a} = 0$

Now since $\color{#856d4d}{a} + \color{#3D99F6}{b} + \color{#BA33D6}{c} = 0 \\ \implies \left(\color{#856d4d}{a} + \color{#3D99F6}{b} + \color{#BA33D6}{c} \right)^2 =0 \\ \implies \color{#856d4d}{a^2} + \color{#3D99F6}{b^2} + \color{#BA33D6}{c^2} + 2 \left(\color{#856d4d}{a} \color{#3D99F6}{b} + \color{#3D99F6}{b} \color{#BA33D6}{c} + \color{#BA33D6}{c} \color{#856d4d}{a} \right) = 0 \\ \implies \color{#856d4d}{a^2} + \color{#3D99F6}{b^2} + \color{#BA33D6}{c^2} + 2 \left(0 \right) = 0 \\ \implies \color{#856d4d}{a^2} + \color{#3D99F6}{b^2} + \color{#BA33D6}{c^2} = 0 \\ \implies \color{#856d4d}{e^{2ix}}+\color{#3D99F6}{e^{2iy}} + \color{#BA33D6}{e^{2iz}} = 0 \\ \implies \left(\color{#856d4d}{\cos(2x)} + \color{#3D99F6}{\cos(2y)} + \color{#BA33D6}{\cos(2z)} \right) + i \left(\color{#856d4d}{\sin(2x)} + \color{#3D99F6}{\sin(2y)} + \color{#BA33D6}{\sin(2z)} \right) = 0 \\ \implies \color{#856d4d}{\cos(2x)} + \color{#3D99F6}{\cos(2y)} + \color{#BA33D6}{\cos(2z)} = 0 \\ \implies \color{#856d4d}{\dfrac{1-\tan^2(x)}{1+ \tan^2(x)}} + \color{#3D99F6}{\dfrac{1-\tan^2(y)}{1+ \tan^2(y)}} + \color{#BA33D6}{\dfrac{1-\tan^2(z)}{1+ \tan^2(z)}} = 0 \\ \implies \sum_{cyc} \left(\color{#856d4d}{\left(1-\tan^2(x) \right)} \color{#3D99F6}{\left(1+\tan^2(y)\right)} \color{#BA33D6}{\left(1+\tan^2(z)\right)} \right) = 0$

So the result $\sum_{cyc} \left( \color{#856d4d}{\left( 1-\tan^2(x) \right)} \color{#3D99F6}{\left( 1+\tan^2(y) \right)} \color{#BA33D6}{\left(1+\tan^2(z)\right)} \right) = 0$ gives our required result to be $\color{#6f00ff}{\boxed{-3}}$ by expanding it.

We get that the question involves complex numbers. Actually, 3 complex numbers. Sum of real parts of all three is equal to sum of complex parts is equal to zero. What we dont get is why the question asks of evaluating an expression involving tan squares. So we bring in three of the simplest complex numbers that satisfy the equation: 1, w, w^2.

Plugging in the values of their tan squares, which are 0, 3, and 3 respectively, we get a solution equal to -3.

Because the question invloves arguments, and angles, and directions. We may generalise the solution by multiplying the roots of unity by e^ix. The symmetry of the system will leave us with the same answer.

I like to simplify where possible. So I started with assuming x=0, and looked for a solution for y and z.

With some thought I determined that z=-y satisfies the sin equation. And I found y=acos(-.5) took care of the cos equation.

Now tan(x)=0, tan(y)=-sqrt(3), tan(z)=sqrt(3) and everything fell out nicely. Most of the terms in the final equation are now 0.

We have 3 + 3 - 9 = -3