How is relativity equation approximated What is the technique with which I can approximate the equation 
$$\frac{mc^2}{\sqrt{1-(\frac{v}{c})^2}}-mc^2$$
when $v\ll c$?
Any hint would be much appreciated
 A: When $v\ll c$, the ratio $\beta = v/c$ is small, so we perform a Taylor expansion about $\beta = 0$;
\begin{align}
  \frac{1}{\sqrt{1-\beta^2}} = (1-\beta^2)^{-1/2} = 1+\frac{1}{2}\beta^2+\frac{3}{8}\beta^4+\cdots
\end{align}
Now plug this into your expression and simplify.
A: This is just a supplement to Josh's correct answer. How to derive the Taylor expansion? Note that the binomial expansion for a power is given by the formula with the combinatorial numbers
$$ (1+x)^n = \sum_{k=0}^\infty \frac{n(n-1)\cdots (n-k+1)}{k!} x^k$$
where the sum could have been stopped at $k=n$ but the infinite upper bound doesn't hurt because the terms with $k\gt n$, are zero for an integer $n$. The $k=0$ term has the factor $1$ in the numerator (no factors in the product).
Now, the formula above may be believed to hold for a fractional $n$, too. For example, for the relevant case $n=-1/2$ (inverted square root), the coefficient is $1$ for $k=0$ and for $k=1$ in the expansion, it is $(-1/2) / 1! = -1/2$. Substituting $x=-\beta^2$, we get $+\beta^2/2$ times $mc^2$ which is $mv^2/2$.
Note that the $k=2$ term in the expansion is $(-1/2)(-3/2)x^2 / 2! = +(3/8)x^2$, also confirming Josh's next term in the expansion.
People with the knowledge of calculus derive the Taylor expansions for particular functions using the $k$-th derivatives evaluated at $x=0$.
