# Prove that $\frac{1}{\sqrt{1-\sin^2{x}}}=\sum\limits_{n=0}^{\infty} \frac{(2n)!(\sqrt{\sin{x}})^{4n}}{4^n (n!)^2}$.

Prove that $$\frac{1}{\sqrt{1-\sin^2{x}}}=\sum_{n=0}^{\infty} \frac{(2n)!(\sqrt{\sin{x}})^{4n}}{4^n (n!)^2}$$

• Do you mean: $$\frac{1}{\sqrt{1-\sin^2{x}}}=\sum_{n=0}^{\infty} \frac{(2n)!(\sqrt{\sin{x}})^{4n}}{4^n (n!)^2}$$ – projectilemotion Apr 1 '17 at 16:33
• Yes👍👍👍☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝👍👍 – creedoos Apr 1 '17 at 16:48
• Ok, the ambiguity of the problem is now settled. What are your thoughts and what have you tried to solve the problem? That way, we don't repeat information you already know. – projectilemotion Apr 1 '17 at 16:53

Hint. One may use the generalized binomial theorem (see here) to get, $$\begin{eqnarray*} \frac{1}{\sqrt{1-u}}=(1-u)^{-1/2}=\sum_{n=0}^\infty \frac{{2n\choose n}}{2^{2n}}u^{n} &, \qquad |u|<1, \end{eqnarray*}$$ then one may put $u:=\sin^2 x$.