`cosec^2alpha=(4x y)/((x+y)^2)[(x , y)!=(0, 0)]` is

sec^2theta=(4x y)/((x+y)^2) is true if and only if x+y!=0 (b) x=y ,x!=0 x=y (d) x!=0,y!=0

The equation cosec theta = (x^(2)+y^(2))/(2 xy) is (x != 0, y != 0)

If tan alpha=(x)/(y) , then show that x "cosec" (alpha)/(3)- y "sec" (alpha)/(3)=3 sqrt(x^(2)+y^(2))(0 lt alpha lt (pi)/(2))

(dy)/(dx) - (y)/(x) + cosec ((y)/(x)) = 0, y = 0 when x = 1.

(dy)/(dx) - (y)/(x) + cosec ((y)/(x)) = 0, y = 0 when x = 1.

Find the particular solution of the differential equation : dy/dx + y cot x = 4x cosec x, (x ≠ 0), given, that y = 0, when x = pi/2

Find the particular solution of the differential equation (dy)/(dx) + y cot x = 4x cosec x (x ne 0) , given that y = 0 , when x = pi/2 .

x=a cosec theta+b cot theta, y=a cosec theta-b cot theta . Then, ((x+y)/(a))^(2)-((x-y)/(b))^(2)=4

If the solution of the differential equation (2x+3y-2)dx+(4x+6y-7)dy=0,y(0)=3 ,is alpha x beta y+3log_(e) abs(2x+3y-gamma)=6 ,then alpha+2 beta+3 gamma is equal to