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?
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?