Expansion formula for $\tan$ We know that we can easily write the expansion of $\sin$ and $\cos$. But writing the expansion of $\tan$ becomes very difficult as coefficients of the powers of $x$ can't be found out easily. Is there a simple way of finding these coefficients, say a general formula for writing up to any number of terms.

If you are only concerned about the answer:
$$tanx=x+\frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\frac{62}{2835}x^9+... $$
As to the method of deriving using the expansions of $sin(x)$ and $cos(x)$, use $$cos(x) \times tan(x) = sin(x)$$
$$(1-\frac{x^2}{2!}+\frac{x^4}{4!}...) \times (a+bx+cx^2+dx^3+ex^4+fx^5...) = (x-\frac{x^3}{3!}+\frac{x^5}{5!}...)$$
You can equate coefficients from both the sides to get some initial constants.
Firstly there are no constants on the right, so $a=0$ (because $a \times 1$ in the expansion of the $LHS$ would give you a constant).
Next, $$bx\times1 = x$$ therefore $b=1$.
Further, $$cx^2\times1 + a\times(\frac{-x^2}{2!}) = 0$$ therefore $c=0$ (because we earlier found $a=0$).
And one more, $$dx^3\times1 + (bx)\times(\frac{-x^2}{2!}) = \frac{-x^3}{3!}$$ Solving this gives you $d = \frac{1}{3}$. (We found $b=1$).
And so on. Do it for as many terms as you feel necessary.  
