Question

of x+y+z=π then prove that cotx/2+coty/2+cotz/2=cotx/2coty/2cotz/2
plz ans. it fast its really urgent.

Answer 1

I've gotten to the point where I've expanded the left hand expression to:

sin(z)cos(x)cos(y)+sin(x)cos(y)cos(z)+sin(y)cos(x)cos(z)sin(x)sin(y)sin(z)sin⁡(z)cos⁡(x)cos⁡(y)+sin⁡(x)cos⁡(y)cos⁡(z)+sin⁡(y)cos⁡(x)cos⁡(z)sin⁡(x)sin⁡(y)sin⁡(z)

I'd imagine that perhaps we're supposed to turn the trig functions into those for a single variable, and I've tried:

sin(z)12(cos(x+y)−sin(x−y))+sin(x)12(cos(y+z)−sin(y−z))+sin(y))12(cos(x+z)−sin(x−z))sin(x)sin(y)sin(z)sin⁡(z)12(cos⁡(x+y)−sin⁡(x−y))+sin⁡(x)12(cos⁡(y+z)−sin⁡(y−z))+sin⁡(y))12(cos⁡(x+z)−sin⁡(x−z))sin⁡(x)sin⁡(y)sin⁡(z)

sin(z)12(cos(π−z)−sin(π−y−2z))+sin(x)12(cos(y+z)−sin(y−z))+sin(y))12(cos(π−y)−sin(π−y−2z))sin(π−y−z)sin(y)sin(z)sin⁡(z)12(cos⁡(π−z)−sin⁡(π−y−2z))+sin⁡(x)12(cos⁡(y+z)−sin⁡(y−z))+sin⁡(y))12(cos⁡(π−y)−sin⁡(π−y−2z))sin⁡(π−y−z)sin⁡(y)sin⁡(z)

But that hasn't worked out for me either