Find the equation of the tangent and normal to the curves.$\;y=\large\frac{1+\sin x}{\cos x}$ at $\;x=\large\frac{\pi}{4}$

Note: This is part 4th of a 4 part question, split as 4 separate questions here.

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) y=\large\frac{1+\sin x}{\cos x} at x=\large\frac{\pi}{4} Step 1: When x=\large\frac{\pi}{4}$$ y=\large\frac{1+\Large\frac{1}{\sqrt 2}}{\Large\frac{1}{\sqrt 2}}$
$\qquad=\sqrt 2+1$
We are required to find the equations of the tangent and the normal at $\bigg(\large\frac{\pi}{4},$$\sqrt 2+1\bigg) Step 2: The slope of the tangent at \bigg(\large\frac{\pi}{4},$$ \sqrt 2+1\bigg)$ is
$m=\large\frac{dy}{dx_{(\Large\frac{\pi}{4},\sqrt 2+1)}}$
$\large\frac{dy}{dx}=\large\frac{\cos x(\cos x)-(1+\sin x)(-\sin x)}{\cos ^2 x}$
$\qquad=\large\frac{\cos^2 x+\sin x+\sin^2 x}{\cos ^2 x}$
$\qquad=\large\frac{1+\sin x}{\cos ^2 x}$
$m=\large\frac{dy}{dx_{(\Large\frac{\pi}{4},\sqrt 2+1)}}$
$\quad=\large\frac{1+\Large \frac{1}{\sqrt 2}}{\Large\frac{1}{2}}$
$\quad=\large\frac{\sqrt 2+1}{\sqrt 2}$$.2 \quad=2 +\sqrt 2 Equation of the tangent at \bigg(\large\frac{\pi}{4},$$ \sqrt 2+1\bigg)$
$y-(\sqrt 2+1)=(2 +\sqrt {2})(x-\large\frac{\pi}{4})$
Step 3:
Normal is $\perp$ to the tangent. Its slope equation of the normal at $\bigg(\large\frac{\pi}{4},$$\sqrt 2+1\bigg) y-(\sqrt 2+1)=\large\frac{-1}{(2 +\sqrt 2)}($$x-\large\frac{\pi}{4})$