Difficult trigonometric problem

[Krishang]

If A, B and C are the angles of triangle ABC, and [imath]\prod \tan \left(\frac{X+Y-Z}{4} \right) = 1[/imath], then find the value of [imath]\frac{\sin(A) + \sin(B) +\sin(C)}{\sin(A)\sin(B)\sin(C)}[/imath](Which is say, [imath]\lambda[/imath]). I am unable to do this question.

I attempted to simplify



[math]\tan\left(\frac{\pi}{4} - \frac{C}{2}\right) \tan\left(\frac{\pi}{4} - \frac{A}{2}\right) \tan\left(\frac{\pi}{4} - \frac{B}{2}\right) = 1 \\ 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right) = 8\lambda\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right) \\ 2\lambda\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right) = 1[/math]
But this didn't simplify in the way which was easy to work with, and I am unsure of any other ways. How to approach this problem?

 

[blamocur]

What are X,Y,Z? What are the components of the product?

 

[Krishang]

What are X,Y,Z? What are the components of the product?

It is notation expressing that X,Y,Z are the angles of the triangle. And the components of the product are taking all 3 angles. Basically,
[math]\tan\left(\frac{A+B-C}{4}\right) \tan\left(\frac{C+B-A}{4}\right) \tan\left(\frac{A+C-B}{4}\right) = 1[/math]

 

[blamocur]

It is notation expressing that X,Y,Z are the angles of the triangle. And the components of the product are taking all 3 angles. Basically,
[math]\tan\left(\frac{A+B-C}{4}\right) \tan\left(\frac{C+B-A}{4}\right) \tan\left(\frac{A+C-B}{4}\right) = 1[/math]

I am having strong doubts about the conditions of this exercise. My quick and dirty script shows that the product can't be equal to 1 for real triangles. Indeed, in one million random triangles it never exceeded 0.01924. It seems that the maximum value is achieved when [imath]A=B=C=\frac{\pi}{3}[/imath] and the product in the exercise is [imath]\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)^3 = \frac{3\sqrt{3}-5}{3\sqrt{3}+5}\approx 0.019238[/imath].

Am I missing something there? Otherwise, could you recheck the problem statement?
Thank you.

 

[Krishang]

I am having strong doubts about the conditions of this exercise. My quick and dirty script shows that the product can't be equal to 1 for real triangles. Indeed, in one million random triangles it never exceeded 0.01924. It seems that the maximum value is achieved when [imath]A=B=C=\frac{\pi}{3}[/imath] and the product in the exercise is [imath]\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)^3 = \frac{3\sqrt{3}-5}{3\sqrt{3}+5}\approx 0.019238[/imath].

Am I missing something there? Otherwise, could you recheck the problem statement?
Thank you.

This is the exact question. You're right, I think. The question is wrong. Even summation wouldn't work over here (I verified using the triple angle identity for tan). Sorry for the trouble and thank you for the insight.