If `cosx+cosy+cosalpha=0` and `sinx + siny + sinalpha=0`, then `cot((x+y)/2)=`

If cos x +cosy+cosalpha=0 , sinx+siny+sinalpha=0 then cot((x+y)/2)=

If cosx + cosy + cosz = 0 and sinx + siny + sinz =0 , then show that cos (x-y) + cos(y - z) + cos(z - x) = - 3/2.

If cosx + cosy + cosz = 0 "and" sinx + siny + sinz =0, "then show that" cos (x-y) + cos(y - z) + cos(z - x) = - 3/2.

If cosx+cosy+cosz=0 and sinx+siny+sinz=0 then show that cos(x-y)+cos(y-z)+cos(z-x)=-3/2

If cosx+cosy+cosz=sinx+siny+sinz=0" then "cos(x-y)= (a) 0 (b) -1/2 (c) 2 (d) 1

if x and y are acute angles such that cosx + cosy = 3/2 and sinx + siny = 3/4 then sin(x+y)=

if x and y are acute angles such that cosx + cosy = 3/2 and sinx + siny = 3/4 then sin(x+y)=

If 1 + cos x cos y + sin x sin y = 0 , then which of the following is / are true 1. cosx + cosy = 0 2. sinx + siny = 0 3. sinx + cosy = 0 chose the correct option from the code given below

If cosx+cosy+cosz=sinx+siny+sinz=0" then "cos(x-y)=

If cosx+cosy+cosz=sinx+siny+sinz=0" then "cos(x-y)=