Let $\left(x_{1}, y_{1}\right)$ be one of the points of contact.
Given curve is $y=\cos x$
$\Rightarrow \frac{ dy }{ dx }=-\sin x$
$\Rightarrow \left|\frac{ dy }{ dx }\right|_{\left( x _{1}, y _{1}\right)}=-\sin x _{1}$
Now the equation of the tangent at $\left( x _{1}, y _{1}\right)$ is
$y - y _{1}\left(\frac{ dy }{ dx }\right)_{\left( x _{1}, y _{1}\right)}\left( x - x _{1}\right)$ $\Rightarrow y - y _{1}=-\sin x _{1}\left(0- x _{1}\right)$
Since, it is given that equation of tangent passes through origin
$\Rightarrow 0-y_{1}=-\sin x_{1}\left(0-x_{1}\right)$
$\therefore y _{1}=- x _{1} \sin x _{1}$ ... (1)
also, point $\left( x _{1}, y _{1}\right)$ lies on
$\therefore y _{1}=\cos x _{1}$
From Eqs (i), (ii) we get
$\sin ^{2} x_{1}+\cos ^{2} x_{1}=\frac{y_{1}^{2}}{x_{1}^{2}}+y_{1}^{2}=1$
$\Rightarrow x_{1}^{2}=y_{1}^{2}+y_{1}^{2} x_{1}^{2}$
Hence, the locus of $x^{2}=y^{2}+y^{2} x^{2}$
$\Rightarrow x^{2} y^{2}=x^{2}-y^{2}$