Answer
$$ y=2\sqrt{2}-2x$$
Work Step by Step
The equation of the tangent line to the graph of $f$ at $x=0$ is:
$$y=f(0)+f'(0)(x-0)$$
$$y=f(0)+f'(0)x$$
So $f(0)=M\tan(0)+N\sec(0)=N$
so:
$$y=N+f'(0)x$$
The first derivative of $f$ is:
$$f'(x)=M\sec^{2}(x)+N\sec(x)\tan(x)$$
$$f'(0)=M\sec^{2}(0)+N\sec(0)\tan(0)=M$$
so:
$$y=N+Mx$$
$$f'(\frac{\pi}{4})=M\sec^{2}(\frac{\pi}{4})+N\sec(\frac{\pi}{4})\tan(\frac{\pi}{4})=2M+\sqrt 2N=0$$
$$f(\frac{\pi}{4})=M\tan(\frac{\pi}{4})+N\sec(\frac{\pi}{4})=M+\sqrt 2N=2$$
The system of equations is:
$$\begin{cases}2M+\sqrt 2N=0\\M+\sqrt 2N=2\end{cases}$$
$$2M+\sqrt 2N-M-\sqrt 2N=0-2$$
$$M=-2$$
so:
$$M+\sqrt 2N=2 \to -2+\sqrt 2N=2 \to N=2\sqrt{2}$$
so the equation of the tangent line is:
$$y=N+Mx \to y=2\sqrt{2}-2x$$
