[y=6],[(x)/(a)+(y)/(b)=a+b],[(x)/(a^(2))quad (xy)/(b_(2))=2quad y=a^(2)]

ab(x^(2)+y^(2))-xy(a^(2)+b^(2))

cos^(-1)(x//a)+cos^(-1)(y//b)=theta ,then (x^(2))/(a^(2))-(2xy)/(ab)cos theta+(y^(2))/(b^(2))=

If (x+1)/(x-1)=(a)/(b) and (1-y)/(1+y)=(b)/(a), then the value of (x-y)/(1+xy) is (2ab)/(a^(2)-b^(2)) (b) (a^(2)-b^(2))/(2ab) (c) (a^(2)+b^(2))/(2ab) (d) (a^(2)-b^(2)backslash)/(ab)

If Cos^(-1)(x/a)+Cos^(-1)(y/b)=theta,"then "x^(2)/a^(2)-(2xy)/(ab)costheta+y^(2)/b^(2)=

Solve :a(x+y)+b(x-y)=a^(2)-ab+b^(2)a(x+y)-b(x-y)=a^(2)+ab+b^(2)

If a=(x)/(x+y) and b=(y)/(x-y), then (ab)/(a+b) is equal to (a) (xy)/(x^(2)+y^(2)) (b) (x^(2)+y^(2))/(xy)( c) (x)/(x+y) (d) ((y)/(x+y))^(2)

The average of the reciprocals of x and y is (a) ((x+y))/((x-y)) (b) ((x+y))/(2xy) (c) (2(x+y))/(xy) (d) (2xy)/((x+y))

The factors of x^(3)-x^(2)y-xy^(2)+y^(3) are (a (x+y)(x^(2)-xy+y^(2))(b)(x+y)(x^(2)+xy+y^(2))(c)(x+y)^(2)(x-y)(d)(x-y)^(2)(x+y)