यदि `cosx+cosy+cosalpha=sinx+siny+sinalpha=0` तब `cot((x+y)/(2))=`

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

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

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

If cosx+cosy+cosalpha=0 and sinx+siny+sinalpha=0 then cot((x+y)/2)= ................. A) sin alpha B) cos alpha C) cot alpha D) sin ((x+y)/(2))

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

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

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

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

Prove that (cosx + cosy)/(sinx - siny) = cot ((x-y)/2)

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