The equation of the curve passing through the point `(1, (pi)/(4))` and having slope of tangent at any point (x, y) as `(y)/(x)-"cos"^(2)(y)/(x)` is equal to

Options

A. `x=e^(1-tan(y//x))`
B. `y= x tan^(-1)(log x//e)`
C. `y=x tan^(-1)log(e//x)`
D. `x=e^(1+cot(y//x))`

Answer

To solve the problem, we need to find the equation of the curve that passes through the point \((1, \frac{\pi}{4})\) and has a slope of the tangent at any point \((x, y)\) given by the expression \(\frac{y}{x} - \cos^2\left(\frac{y}{x}\right)\).

### Step 1: Set up the differential equation
We start with the given slope of the tangent:
\[ \frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right) \]

### Step 2: Use substitution
Let \(v = \frac{y}{x}\), which implies \(y = vx\). We can differentiate this with respect to \(x\):
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Now, substituting \(y = vx\) into the differential equation gives:
\[ v + x\frac{dv}{dx} = v - \cos^2(v) \]

### Step 3: Simplify the equation
We can simplify this equation:
\[ x\frac{dv}{dx} = -\cos^2(v) \]

### Step 4: Separate variables
Rearranging gives:
\[ \frac{dv}{-\cos^2(v)} = \frac{dx}{x} \]

### Step 5: Integrate both sides
Now we integrate both sides:
\[ \int \frac{dv}{-\cos^2(v)} = \int \frac{dx}{x} \]
The left side integrates to \(-\tan(v)\) and the right side integrates to \(\ln|x| + C\):
\[ -\tan(v) = \ln|x| + C \]

### Step 6: Substitute back for \(v\)
Substituting back \(v = \frac{y}{x}\):
\[ -\tan\left(\frac{y}{x}\right) = \ln|x| + C \]

### Step 7: Use the initial condition
We know the curve passes through the point \((1, \frac{\pi}{4})\). Substituting \(x = 1\) and \(y = \frac{\pi}{4}\):
\[ -\tan\left(\frac{\pi}{4}\right) = \ln(1) + C \]
Since \(\tan\left(\frac{\pi}{4}\right) = 1\) and \(\ln(1) = 0\), we have:
\[ -1 = 0 + C \implies C = -1 \]

### Step 8: Write the final equation
Substituting \(C\) back into the equation gives:
\[ -\tan\left(\frac{y}{x}\right) = \ln|x| - 1 \]
or rearranging:
\[ \tan\left(\frac{y}{x}\right) = 1 - \ln|x| \]

### Final Result
Thus, the equation of the curve is:
\[ \tan\left(\frac{y}{x}\right) = 1 - \ln|x| \]