Question

If u = tan-'(), then​

Answer 1

Answer:u = tan-1 (x3+y3)/(x-y)

tan u = (x3+y3)/(x-y)

Let tan u = z …(i)

So z = (x3+y3)/(x-y)

= x3(1+y3/x3)/x(1-y/x)

= x2(1+y3/x3)/(1-y/x)

= x2(1+(y/x)3)/(1-y/x)

Here z is a homogeneous function of the form xn f(y/x).

Here n = 2

So by Euler’s theorem

x∂z/∂x + y ∂z/∂y = nz …(ii)

∂z/∂x = sec2 u ∂u/∂x (from (i))

∂z/∂y = sec2 u ∂u/∂y

Substitute above 2 equations in (ii)

x sec2 u ∂u/∂x + y sec2 u ∂u/∂y = 2 tan u

x ∂u/∂x + y ∂u/∂y = 2 tan u/sec2 u

x ∂u/∂x + y ∂u/∂y = 2 sin u cos u

x ∂u/∂x + y ∂u/∂y = sin 2u

Hence option (2) is the answer.

Step-by-step explanation: please mark me as brainiest

Answer 2

Answer:

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