`y = cos x + C : y'+sin x = 0`

Solution of differential equation (2x cos y + y ^ (2) cos x) dx (2y sin xx ^ (2) sin y) dx = 0 is (a) x ^ (2) cos y + y ^ (2) sin x = C (b) x cos yy sin x = C (c) x ^ (2) cos ^ (2) y + y ^ (2) sin ^ (2) x = C (d) x cos y + y sin x = C

The solution of (dy) / (dx) + (y cos x + sin y + y) / (sin x + x cos y + x) = 0 is

Solve(cos y + y cos x ) dx + sin (x - x sin y) dy = 0

What ( sin x cos y + cos x sin y) (sin x cos y - cos x sin y) equal to ?

If cos(x - y)= -1 , show that cos x + cos y = 0 and sin x + sin y = 0 .

cos x + cos y + cos z = 0 and sin x + sin y + sin z, then cos ^(2) (( x-y)/( 2)) =

Prove that : (sin (x+y))/(sin (x-y) )= (sin x. cos y + cos x . Sin y)/(sin x. cos y-cos x. sin y)

Prove that : (sin (x+y))/(sin (x-y) )= (sin x. cos y + cos x . Sin y)/(sin x. cos y-cos x. sin y)

If sin x + sin y + sin z = 0 and cos x + cos y + cos z = 0 then prove that cot (x+y)/2 = cot z and cos (x-y)/2 = pm 1/2

cos x + cos y + cos z = 0 and sin x + sin y + sin, then cos ^ (2) ((xy) / (2)) =