Trigonometry $\tan \theta = \dfrac43$ If $\tan \theta = \dfrac43$, find $\cos \theta$ and $\sin \theta$.
I have the answers, but I don't know what to do at all, is it going to be easy?

Geometric Approach
Make a (rough) sketch of a right triangle with one angle $\theta$. Then the opposite side and adjacent side are in the ratio $4:3$, as $\tan\theta = \dfrac{\text{Opposite side}}{\text{Adjacent side}}$. You can assume the sides to be exactly $4$ and $3$. Calculate the hypotenuse using Pythagorean theorem. Then calculate $\sin \theta$ and $\cos \theta$.
As 5xum points out, this is not very general. A more general approach is the algebraic one.

Algebraic Approach
$
\tan\theta = \dfrac{4}{3} \Rightarrow\\
\dfrac{\sin\theta}{\cos\theta} = \dfrac{4}{3} \Rightarrow\\
\cos\theta = \dfrac{3}{4}\sin\theta
$
Now,
$
\cos^2\theta + \sin^2\theta = 1 \Rightarrow\\
\dfrac{9}{16}\sin^2\theta + \sin^2\theta = 1 \Rightarrow\\
\sin^2\theta = \dfrac{16}{25} \Rightarrow\\
\sin\theta = \pm \dfrac{4}{5}
$
Then $\cos\theta = \dfrac{\sin\theta}{\tan\theta} = \pm\dfrac{3}{5}$ (with respective signs).
Thus
$$\boxed{
\begin{matrix}
\sin\theta = \dfrac{4}{5},\ \cos\theta = \dfrac{3}{5}\\
\text{or}\\
\sin\theta = -\dfrac{4}{5},\ \cos\theta = -\dfrac{3}{5}
\end{matrix}
}
$$
