If cot(x+y)=0, then write the value of sin(x+2y).Show the solution

HERE , cot(x+y)=0

SO, ON SOLVING WE GET ,TANxTANy=1.

so, SINxSINy=COSxCOSy

NOW , SIN(x+2y)=SINxCOS2y+COSxSIN2y

=SINx(1-2SINySINy)+COSx2SINyCOSy

=SINx-2SINySINySINx+2SINySINySINx

=SINx

From the given equation,
cot(x+y)=0
so, cos(x+y)/sin(x+y)=0
=> cos(x+y)=0
=> x+y= ∏/2
so, y= ∏/2- x
Now sin(x+2y)= sin(x+∏-2x)
= sin(∏-x)
= sin x
Hence, sinx is the answer..

cot(a+b)=0 then,cos(a+b) =0 and,a+b=npi+pi/2 a=npi+pi/2-b and,b=npi+pi/2-a sin(a+2b)=sin(npi+pi/2+b)=+cos or -cos sin(a+2b)=sin(2npi+pi/2-a)=sina