The solution of $(\frac{xdx + ydy}{xdy - ydx}) = \sqrt{(\frac{a^{2} - x^{2} - y^{2}}{x^{2} + y^{2}})}$ is

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$\sqrt{(x^{2} + y^{2})} = asin \{(\tan^{- 1} y / x) + cons \tan t\}$

$\sqrt{(x^{2} + y^{2})} = a \{((\sin^{- 1} y / x) + cons \tan t)\}$

$\sqrt{(x^{2} + y^{2})} = a \{\tan (\cos^{- 1} y / x) + cons \tan t\}$

$\sqrt{(x^{2} + y^{2})} = acos \{(\tan^{- 1} y / x) + cons \tan t\}$

answer is A.

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Detailed Solution

Taking $x = rcos θ & y = rsin θ so that$

$x^{2} + y^{2} = r^{2} & \frac{y}{x} = \tan θ$

We have $xdx + ydy = r \cdot dr and xdy - ydx = x^{2} se c^{2} θdθ = r^{2} dθ$

The given equation can be transformed into $\frac{rdr}{r^{2} dθ} = \sqrt{(\frac{a^{2} - r^{2}}{r^{2}})}$

$\Rightarrow \frac{dr}{dθ} = \sqrt{(a^{2} - r^{2})} \Rightarrow \frac{dr}{\sqrt{(a^{2} - r^{2})}} = dθ$

Integrating both sides then we get

$\sin^{- 1} (\frac{r}{a}) = θ + C$

$\Rightarrow \sqrt{x^{2} + y^{2}} = a \sin [\tan^{- 1} (y / x) + \cos \tan t]$

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