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Formation of a Differential Equation from a General Solution

If y=cot-1√...

Question

If $y = c o t^{- 1} \sqrt{\frac{1 - s i n x}{1 + s i n x}},$ find $\frac{d x}{d y}$ .

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Solution

$y = {cot}^{- 1} \sqrt{\frac{1 - sin x}{1 + sin x}} = {cot}^{- 1} \sqrt{\frac{1 + cos (\frac{π}{2} + x)}{1 - cos (\frac{π}{2} + x)}}$
$y = {cot}^{- 1} {cot (\frac{π}{4} + \frac{x}{2})} = \frac{π}{4} + \frac{x}{2}$
$∴ \frac{d y}{d x} = \frac{1}{2}$

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y = {cot}^{- 1} \frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}

\frac{d y}{d x}

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y = {cot}^{- 1} [\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}]

\frac{d y}{d x} =

Y = {cot}^{- 1} [\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}]

\frac{d y}{d x}

is independent of

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y = \frac{sin x}{1 + \frac{cos x}{1 + \frac{sin x}{1 + cos x . . . \infty}}}

\frac{d y}{d x} = ?

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