I'm trying to find the value of this question :

Cos β / (sinα.sinγ) + cos α / (sin β. sin γ)+ cos γ/ (sin α. Sin β)

Provided that α, β, γ is the angle in a triangle

Please help me to do this

The answer in the book says that the value is 2, but how on earth are we going to find "2"

I already tried like this

= Cos β / (sinα.sinγ) + cos α / (sin β. sin γ)+ cos γ/ (sin α. Sin β)

= cos β sin β + sin α cos α + sin γ cos γ / sin α sin β sin γ

= sin 2β + sin 2α + sin 2 γ/ 2 (sin α sin β sin γ)

Using identities sin A + sin B = 2 sin (A+B/2) cos (A-B/2)

And 2 sin A sin B =sin (A+B) – sin (A-B)

We have

= 2 sin β cos 0 + 2 sin α cos 0 + 2 sin γ cos 0 / (sin ( α+β ) – sin ( α- β)) sin γ

= 2 ( sin α + sin β + sin γ) / (sin γ( 2 sin β cos α))

Then 2 sin β sin γ becomes 2 sin γ cos β

= 2 (sin α + sin β + sin γ) / 2 cos α cos β sin γ

= sin α + sin β + sin γ / cos α cos β sin γ

= ( (sin α + sin β / sin γ) x 1/(cos α cos β) + 1/ cos α cos β

= 1/cos α cos β((sin α + sin β / sin γ) +1)

2cos A cos B = cos (A+B) + cos (A-B)

= (2 / (cos (α + β) + cos (α- β) ) ((sin α + sin β / sin γ) +1)

Please help me ! thanks