# If α + β + γ = π  and tan(β+γ-α)/4 . tan(γ+α -β)/4 .tan(α+β-γ)/4 = 1. Prove that 1 + cos α + cos β + cos γ = 0

Q: If α + β + γ = π  and $tan\frac{(β + γ-α)}{4} . tan\frac{(γ + α -β)}{4} .tan\frac{(α + β-γ)}{4} = 1$ . Prove that 1 + cos α + cos β + cos γ = 0 .

Solution : Let $A = tan\frac{(β + γ-α)}{4}$ , $B = tan\frac{(γ + α -β)}{4}$ and $C = tan\frac{(α + β-γ)}{4}$

Therefore , $tanA . tanB . tanC = 1$

$tanA . tanB = \frac{1}{tanC}$

$\large \frac{sinA}{cosA} . \frac{sinB}{cosB} = \frac{1}{tanC}$

$\large \frac{sinA sinB -cosA cosB}{sinA sinB + cosA cosB} = \frac{1-tanC}{1+tanC}$

$\large – \frac{cos(A+B)}{cos(A-B)} = \frac{sin(\frac{\pi}{4}-C)}{cos(\frac{\pi}{4}-C)}$

$\large 2 sin(\frac{\pi}{4}-C) cos(A-B) + 2 cos(\frac{\pi}{4}-C) cos(A+B) = 0$

$sin(\frac{\pi}{4}-C + A-B) + sin(\frac{\pi}{4}-C – A+B) + cos(\frac{\pi}{4}-C + A+B) + cos(\frac{\pi}{4}-C – A-B) = 0$ …(1)

Since $A-B-C = \frac{\pi}{4}-\alpha$

Similarly , $B-C -A = \frac{\pi}{4}-\beta$

and , $C – A – B = \frac{\pi}{4}-\gamma$

and $C + A + B = \frac{\alpha + \beta + \gamma}{4} = \frac{\pi}{4}$

Equation (1) reduces to

$\large sin(\frac{\pi}{4} + \frac{\pi}{4} – \alpha) + sin(\frac{\pi}{4} + \frac{\pi}{4} – \beta) + cos(\frac{\pi}{4} – \frac{\pi}{4} + \gamma) + cos(\frac{\pi}{4} – \frac{\pi}{4} ) = 0$

$\large cos\alpha + cos\beta + cos\gamma + 1 = 0$