To solve the given differential equation:
\(\frac{dy}{dx} = \sin(x + y) \tan(x + y) - 1\)
we first rewrite it in a form that is easier to integrate. Observe that:
\(\tan(x + y) = \frac{\sin(x + y)}{\cos(x + y)}\)
Hence, the equation becomes:
\(\frac{dy}{dx} = \frac{\sin^2(x + y)}{\cos(x + y)} - 1\)
We recognize that \(\sin^2(x + y) = 1 - \cos^2(x + y)\), so we substitute:
\(\frac{\sin^2(x + y)}{\cos(x + y)} - 1 = \frac{1 - \cos^2(x + y)}{\cos(x + y)} - 1\)
Simplifying, we get:
\(\frac{dy}{dx} = \frac{1}{\cos(x + y)} - \cos(x + y) - 1\)
At this point, multiply both sides by dx to separate variables:
\(dy = \left(\frac{1}{\cos(x+y)} - \cos(x+y) - 1\right) dx\)
Integrate both sides. The first term integrates to give \(\int \sec(x+y) dx = \ln |\sec(x+y) + \tan(x+y)|\) and the second term integrates directly using standard integrals.
This leads ultimately to:
\(x + \cosec(x + y) = C\)
where \(C\) is the constant of integration.\
Therefore, the correct answer is:
\(x + \cosec(x + y) = C\)