let's take the postive value for a while (that is the value in first quadrant.)
[tex] \tan(x) =\frac{\text{perpendicular or opposite side}}{\text{base or adjacent side}} \\ \tan(x)=2 \\ \text{if you imagine a triangle with perpendicular =2 and base =1, and use Pythagoras theorem, you will get hypotenuse} \\ h^2= 2^2+1^2 \\ h^2=4+1 =5 \\ h=\sqrt5 \\ \cos(x)=\frac{\text{base or adjacent side}}{\text{hypotenuse}} \\ \cos(x)=\frac{1}{\sqrt5} \\ \text{ since x is in second quadrant, where both tangent and cosine are negative, just add a negative sign} \\ \cos(x)= -\frac{1}{\sqrt5} [/tex]
