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Question

Solution of $cos invisible function application x fraction numerator d y over denominator d x end fraction plus y sin invisible function application x equals 1$ is

$y \sec x \tan x = c$
$y \sec x \tan x = c$
$y \tan x = \sec x + c$
$y \tan x = \sec x \tan x + c$

The correct answer is: $y sec invisible function application x tan invisible function application x equals c$

Given equation can be written as $fraction numerator d y over denominator d x end fraction plus y tan invisible function application x equals sec invisible function application x$
$therefore$ I.F. $equals e to the power of not stretchy integral tan invisible function application x d x end exponent equals e to the power of logsec invisible function application x end exponent equals sec invisible function application x$
Hence solution is $y sec invisible function application x equals not stretchy integral sec to the power of 2 end exponent invisible function application x plus c equals tan invisible function application x plus c$ .

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