solve : dy/dx= cos (x+y)
dy/dx=cos(x+y) equn(1.)
z=x+y. equn(2.)
diff. w.r.t. x , we get,
dz/dx = 1 + dy/dx.
dz/dx = 1+ cos(x+y) from equn(1.)
dz/dx = 1+ cos z from equn(2.)
dz/(1+cosz) = dx.
dz/(2 cos^2(z/2)) = dx.
so, sec^2(z/2) dz / 2 =dx.
so, sec^2(z/2) dz = 2dx.
integarting both sides, we get
tan(z/2) / (1/2) = 2x.
so, 2tan(z/2) =2x.
so, tan(z/2) = x.
so, tan((x+y)/ 2) =x is ans.
z=x+y. equn(2.)
diff. w.r.t. x , we get,
dz/dx = 1 + dy/dx.
dz/dx = 1+ cos(x+y) from equn(1.)
dz/dx = 1+ cos z from equn(2.)
dz/(1+cosz) = dx.
dz/(2 cos^2(z/2)) = dx.
so, sec^2(z/2) dz / 2 =dx.
so, sec^2(z/2) dz = 2dx.
integarting both sides, we get
tan(z/2) / (1/2) = 2x.
so, 2tan(z/2) =2x.
so, tan(z/2) = x.
so, tan((x+y)/ 2) =x is ans.