If `s e c^2theta=(4x y)/((x+y)^2)` is true if and only if (a)`x+y!=0` (b) `x=y , x!=0` (c) `x=y` (d) `x!=0,y!=0`

sin^2 theta= (4xy)/(x+y)^2 is true if and only if (A) x-y!=0 (B) x=-y (C) x+y!=0 (D) x!=0,y!=0

Prove that sin^2theta=(x+y)^2/(4xy) is possible for real values of x and y only when x = y , y!=0 and x!=0 .

The equations of tangents to the circle x^2+y^2-6x-6y+9=0 drawn from the origin in (a). x=0 (b) x=y (c) y=0 (d) x+y=0

x=2, y=−1 is a solution of the linear equation (a) x+2y=0 (b) x+2y=4 (c) 2x+y=0 (d) 2x+y=5

Statement I: sec^2theta=(4x y)/((x+y)^2) is positive for all real values of x\ a n d\ y only when x=y because Statement II: t^2geq0AAt in R\ a. A b. \ B c. \ C d. D

The point of intersection of the tangents of the parabola y^2=4x drawn at the endpoints of the chord x+y=2 lies on (a) x-2y=0 (b) x+2y=0 (c) y-x=0 (d) x+y=0

The equation of tangent to hyperbola 4x^2-5y^2=20 which is parallel to x-y=2 is (a) x-y+3=0 (b) x-y+1=0 (c) x-y=0 (d) x-y-3=0

The equation of the normal to the curve x=a\ cos^3theta,\ \ y=a\ sin^3theta at the point theta=pi//4 is (a) x=0 (b) y=0 (c) x=y (d) x+y=a

Let L be a normal to the parabola y^2=4x dot If L passes through the point (9, 6), then L is given by (a) y-x+3=0 (b) y+3x-33=0 (c) y+x-15=0 (d) y-2x+12=0

If A=[(5,x),( y,0)] and A=A^T , then a) x=0,\ \ y=5 (b) x+y=5 (c) x=y (d) none of these