The equation of the curve through $(0, π / 4)$ satisfying the differential equation $e^{x} t a n y d x + (1 + e^{x}) s e c^{2} y d y = 0$ is given by

(e^x + 1)

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(e^x - 1) tan y = 0

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(e^x + 1) tan y = 2

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(e^x + 1) cot y = 2

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Solution

The correct option is C
(e^x + 1) tan y = 2

$\frac{e^{x}}{e^{x} + 1} d x + \frac{s e c^{2} y d y}{t a n y} = 0$

$(e^{x} + 1) t a n y = c)$
$x = 0 \Rightarrow y = \frac{π}{4}$
$c = 2$
$(e^{x} + 1) t a n y = 2$

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