The degree of the differential equation satisfying the relation $\sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} = λ (x \sqrt{1 + y^{2}} - y \sqrt{1 + x^{2}})$ is

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Solution

The correct option is A 1
On Putting $x = t a n A, y = t a n B$ we get
$s e c A + s e c B = λ (t a n A s e c B - t a n B s e c A) c o s A + c o s B = λ (s i n A - s i n B) t a n (\frac{A - B}{2}) = \frac{1}{λ} t a n^{- 1} x - t a n^{- 1} y = 2 t a n^{- 1} \frac{1}{λ}$
On differentiating $\frac{1}{1 + x^{2}} - \frac{1}{1 + y^{2}} \frac{d y}{d x} = 0$

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Similar questions

The degree of the differential equation satisfying the relation

$\sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} = λ (x \sqrt{1 + y^{2}} - y \sqrt{1 + x^{2}})$ is

\sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} = λ (x \sqrt{1 + y^{2}}) - y \sqrt{1} + x^{2})

\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y)

e^{y \sqrt{1 - x^{2}} + x \sqrt{1 - y^{2}}} = c e^{x}

\sqrt{1 - x^{2}} + \sqrt{1 + y^{2}} = a (x - y)

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