The degree of the differential equation satisfying the relation `sqrt(1+x^2) + sqrt(1+y^2) = lambda (x sqrt(1+y^2)- ysqrt(1+x^2))` is

Options

A. 1
B. 2
C. 3
D. None of these

Answer

a

Explanation

On putting x = tan A and y = tan B in the given relation ,
we get ` cos A + cos B = lambda ( sin A - sinB)`
` rArr tan.((A-B)/2) = 1/lambda `
` rArr tan^(-1) x - tan^(-1) y = 2 tan^(-1) (1/lambda )`
On differentiating w.r.t x, we get
`1/(1+x^(2)) - 1/(1+y^(2)) = 0 `
` rArr (dy)/(dx) = (1+y^(2))/(1+x^(2))`
Clearly , it is a differential equation of degree `