We can use the definition of Big-O notation to simply prove that
$\sin(z)=z-\frac{z^3}{6}+O(z^5)$ as $z\rightarrow 0$,
$\cos(z)=1-\frac{z^2}{2}+O(z^4)$ as $z\rightarrow 0$ and
$\tan(z)=z+\frac{z^3}{3}+O(z^5)$ as $z\rightarrow 0$.
by using the power series of sin, cos and tan.
But my question is we know that $\tan(z)=\frac{\sin(z)}{\cos(z)}$, so we can write
$\tan(z)=\frac{z-\frac{z^3}{6}+O(z^5)}{1-\frac{z^2}{2}+O(z^4)}$ by the previous results.
However, how to use the algebra of Big-O to show that
$z+\frac{z^3}{3}+O(z^5)=\frac{z-\frac{z^3}{6}+O(z^5)}{1-\frac{z^2}{2}+O(z^4)}$?
Thanks in advance.
You mistyped your $\cos$, that should be $\cos z = 1 - \frac{z^2}{2} + O(z^4)$, and in your fraction for $\tan$, you forgot the leading terms of numerator and denominator (and have the wrong signs for the next term in each). Correcting that, expanding the denominator into (the beginning of) a geometric series yields
$$\begin{align} \tan z &= \frac{z - \frac{z^3}{6} + O(z^5)}{1-\frac{z^2}{2} + O(z^4)}\\ &= \left(z - \frac{z^3}{6} + O(z^5)\right)\cdot \left(1 + \left(\frac{z^2}{2} + O(z^4)\right) + O\left(\left(\frac{z^2}{2} + O(z^4)\right)^2\right)\right)\\ &= \left(z - \frac{z^3}{6} + O(z^5)\right)\cdot \left(1 + \frac{z^2}{2} + O(z^4)\right)\\ &= z - \frac{z^3}{6} + O(z^5) + \frac{z^3}{2} - \frac{z^5}{12} + O(z^7) + z\cdot O(z^4)\\ &= z + \left(\frac{1}{2} - \frac{1}{6}\right)z^3 + O(z^5)\\ &= z + \frac{z^3}{3} + O(z^5). \end{align}$$