Given, equation of parabola is $y^{2}=4 x$.
Here, $a=1$
Now, equation of tangent to the parabola in slope form is
$y=m x+\frac{a}{m} $
$ \Rightarrow y=m x+\frac{1}{m}\,\,\,\,\,\,\,...(i)$
Also given that tangent to the parabola inclined at an angle of $\frac{\pi}{4}$ to the (+ve) direction of $x$ -axis.
$\therefore m=\tan \frac{\pi}{4}=1 $
Then, $y=(1) x+1 \,\,\,\,\,\,$ [from Eq.(i)]
$\Rightarrow x-y+1=0$