[(x+y)/(6)=4(3 pi)/(15)=(x-y)/(3)-(19)/(4)],[" Substitution rethod "]

Solve (x)/(6)+(y)/(15)=4 , (x)/(3)-(y)/(12)=(19)/(4)

{:((x)/(3) + (y)/(15) = 4),((x)/(3) - (y)/(12) = (19)/(4)):}

(x-4)(x-3)=(x-6)(x-5) (y-9)(y-3)= (y-4)(y-3)

Draw the graph of y= sqrt(2) sin (x+ (pi)/(4)) for the values of x= 0 , (pi)/(4) , (pi)/(2) ,(3pi)/(4) in [(pi)/(4) ,(3pi)/(4)] and find the maximum value of y.

Let y(x) be the solution of the differential equation (dy)/(dx)+(3y)/(cos^(2)x)=(1)/(cos^(2)x) and y((pi)/(4))=(4)/(3) then vaue of y(-(pi)/(4)) is equal to (a)-(4)/(3)(b)(1)/(3)( c) e^(6)+(1)/(3)(d)3

Prove that tan^(-1)((x)/(y))-tan^(-1)((x-y)/(x+y)) is (pi)/(4) and Not(-3 pi)/(4)

tan^(-1)((x)/(y))-tan^(-1)((x-y)/(x+y)) is (A) (pi)/(2)(B)(pi)/(3)(C)(pi)/(4)(D)(pi)/(4) or 3(pi)/(4) is (A) (pi)/(2)(B)

The area of the region given by x^(2)+y^(2)-6y<=0&y<=x is (9 pi-9)/(4)b(9 pi-18)/(4)c.(3 pi+9)/(4)d .none of these