`tan y (dy)/(dx) = sin x + cos x`

Answer

To solve the differential equation \( \tan y \frac{dy}{dx} = \sin x + \cos x \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the given equation:
\[ \tan y \frac{dy}{dx} = \sin x + \cos x \]
We can rearrange this to separate the variables:
\[ \tan y \, dy = (\sin x + \cos x) \, dx \]

### Step 2: Integrate both sides
Now we will integrate both sides:
\[ \int \tan y \, dy = \int (\sin x + \cos x) \, dx \]

### Step 3: Solve the left-hand side
The integral of \( \tan y \) is:
\[ \int \tan y \, dy = -\ln |\cos y| + C_1 \]

### Step 4: Solve the right-hand side
Now, we integrate the right-hand side:
\[ \int (\sin x + \cos x) \, dx = \int \sin x \, dx + \int \cos x \, dx \]
The integrals are:
\[ \int \sin x \, dx = -\cos x + C_2 \quad \text{and} \quad \int \cos x \, dx = \sin x + C_2 \]
Combining these gives:
\[ -\cos x + \sin x + C_2 \]

### Step 5: Combine the results
Now we can set the results of the integrals equal to each other:
\[ -\ln |\cos y| = -\cos x + \sin x + C \]
where \( C = C_2 - C_1 \).

### Step 6: Simplify the equation
To simplify, we can multiply through by -1:
\[ \ln |\cos y| = \cos x - \sin x - C \]

### Step 7: Exponentiate both sides
Exponentiating both sides gives:
\[ |\cos y| = e^{\cos x - \sin x - C} = \frac{e^{\cos x - \sin x}}{e^C} \]
Let \( K = e^{-C} \), then:
\[ |\cos y| = K e^{\cos x - \sin x} \]

### Final Result
Thus, the solution to the differential equation is:
\[ \cos y = \pm K e^{\cos x - \sin x} \]