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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

  1. y sec invisible function application x tan invisible function application x equals c    
  2. y sec invisible function application x tan invisible function application x equals c    
  3. y tan invisible function application x equals sec invisible function application x plus c    
  4. y tan invisible function application x equals sec invisible function application x tan invisible function application x plus 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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