Pi is confusing me deeply and I really need help... A short explaination about how to use Pi would be greatly appreciated!
for the tangent you need a line of the form y = mx+c
m is the gradient of tan(x) at the point, pi/4.. this can be found by differentiating tan(x) with respect to x
then to find c, you have a point on the line so substitute into your equation and you can find c.
The gradient of the normal is -1/m, where m is the gradient of teh tangent
The number $\displaystyle \pi$ is confusing you? It is about 3.141592... but is irrational and its decimal expansion continues infinitely. Historically, the first use of $\displaystyle \pi$ was as the ratio of a circle to a diameter. From that we can eventually get things like $\displaystyle sin(\pi)= 0$, $\displaystyle sin(\pi/4)= \sqrt{2}/2$, $\displaystyle cos(\pi/4)= \sqrt{2}/2$, $\displaystyle tan(\pi/4)= 1$.
I was a bit surprised, on opening a thread titled "Finding the Tangent and normal line", to find a question about the number $\displaystyle \pi$. Generally one learns about trig functions and their values, especially at integer fractions of $\displaystyle \pi$ well before a Calculus class.
The derivative of tan(x) is $\displaystyle sec^2(x)=\frac{1}{cos^2(x)}$. So the derivative, and slope of the tangent line, of tan(x) at $\displaystyle x= \pi/4$ is $\displaystyle sec^2(\pi/4)= \frac{1}{cos^2(\pi/4)}= \left(\frac{2}{\sqrt{2}}\right)^2= 2$. The tangent line to tan(x) at $\displaystyle x= \pi/4$ has slope 2 and passes through the point $\displaystyle (\pi/4, tan(\pi/4)= (\pi/4, 1)$. The normal line, perpendicular to the tangent line, has slope -1/2 and passes through the point $\displaystyle (\pi/4, 1)$.
By definition, the derivative of a function at a point is the slope of the tangent line at that point. HallsofIvy just showed you how to obtain the derivative, then how to plug in the point to find the value of the slope of the tangent line at that point. What is troubling you?
Here is some algebra you should know: $\left( \dfrac{a}{b} \right)^c = \dfrac{a^c}{b^c}$
Applying that: $\left( \dfrac{2}{\sqrt{2}} \right)^2 = \dfrac{2^2}{(\sqrt{2})^2} = \dfrac{4}{2} = 2 \neq 1$