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Q.
If the gradient of the tangent at any point $(x, y)$ of a curve which passes through the point $\left(1, \frac{\pi}{4}\right)$ is $\left\{\frac{y}{x}- sin^{2} \left(\frac{y}{x}\right)\right\}$, then equation of the curve is
Given, $\frac{d y}{d x}=\frac{y}{x}-\sin ^{2}\left(\frac{y}{x}\right)$
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$v+x \frac{d v}{d x} =v-\sin ^{2} v $
$ \Rightarrow -\operatorname{cosec}^{2} v d v=\frac{d x}{x} $
Integrating both sides, we get $-\int \operatorname{cosec}^{2} v d v=\int \frac{d x}{x}$
$ \Rightarrow \cot v =\log x+C $
$ \cot \frac{y}{x}=\log x+C $
This curve passes through the ponit $\left(1, \frac{\pi}{4}\right)$
$\therefore C=1 $
$ \Rightarrow \cot \frac{y}{x}=\log x+\log _{e} e $
$\Rightarrow \cot \frac{y}{x}=\log x e $
$ \Rightarrow y=x \cot ^{-1}(\log x e)$