The slope of the tangent at \((x, y)\) to a curve passing through \(\left(1, \frac{\pi}{4}\right)\) is given by \(\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right)\), then the equation of the curve is

According to the given condition,
\(
\frac{d y}{d x}=\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right)
\)
On putting \(y=v x\)
\(
\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}, \text { we get }
\)
\(
\begin{aligned}
& v+x \frac{d v}{d x}=v-\cos ^{2} v \\
\Rightarrow & \frac{d v}{\cos ^{2} v}=-\frac{d x}{x} \\
\Rightarrow & \sec ^{2} v d v=\frac{-1}{x} d x
\end{aligned}
\)
On integrating both sides, we get
\(
\tan v=-\log x+\log c
\)
\(
\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log c
\)
Since, this curve is passing through \((1, \pi / 4)\).
\(
\begin{array}{l}
\therefore \tan \left(\frac{\pi}{4}\right)=-\log 1+\log c \Rightarrow \log c=1 \\
\therefore \tan \left(\frac{y}{x}\right)=-\log x+1 \\
\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log e \\
\Rightarrow y=x \tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]
\end{array}
\)