The solution of the differential equation $$dydx=$$$$sin$$⁡(x+y)$$tan$$⁡(x+y)−$$1$$ is

Correct option is 2x+cosec(x+y)=c

Questions

$The solution of the differential equation \frac{dy}{dx} = \sin (x + y) \tan (x + y) - 1 is$

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cosec(x+y)+tan(x+y)=x+c

x+cosec(x+y)=c

x+tan(x+y)=c

x+sec(x+y)=c

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detailed solution

Correct option is B

Put x+y=t→1+dydx=dtdx⇒∫cosectcott⁡dt=∫dx⇒-cosec⁡(x+y)+c=x

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The solution of the differential equation dydx=sin⁡(x+y)tan⁡(x+y)−1 is

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$The solution of the differential equation \frac{dy}{dx} = \sin (x + y) \tan (x + y) - 1 is$