`y=A cosx^2+B sin x^2` से अवकल समीकरण बनाइए |

y = (cosx + sin x)/(cosx - sin x), prove that dy/dx = sec^2(pi/4 + x)

What is the value of [(sin x + sin y)(sin x - sin y)]//[(cosx + cosy)(cosy- cosx)] ? [(sin x + sin y)(sin x - sin y)]//[(cosx + cosy)(cosy- cosx)] का मान क्या है?

The area bounded by the curve y=cosx and y=sin 2x, AA x in [(pi)/(6), (pi)/(2)] is equal to

The area bounded by the curve y=cosx and y=sin 2x, AA x in [(pi)/(6), (pi)/(2)] is equal to

The area enclosed by the curves y= sinx+cosx and y = | cosx-sin x | over the interval [0,pi/2]

sin 2x+cosx=0

If y = (log_(cosx) sin x)(log_(sinx) cosx)^-1 + sin^-1((2x)/(1+x^2)) , find dy/dx at x = pi/4

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)=