A Problem on Beta distribution . 
In this problem i know that $X\sim B(m,n)$ and $(1-X)\sim B(n,m)$
After putting values in $Y_i$ i got this $\dfrac{x^2}{1-x^2}$  I have not idea what do further . I am confused.   
 A: Hint: Find the MGF of $Y_i$ (after you have the density) and then use it to compute the distribution of $U_n$ then $\alpha$ should be the standard deviation of the resulting distribution by the central limit theorem. 
Edited: You should find this useful:
The following variance of the variable X divided by its mirror-image ($X/(1−X)$ results in the variance of the "inverted beta distribution" or beta prime distribution (also known as beta distribution of the second kind or Pearson's Type VI):
${\displaystyle \operatorname {var} \left[{\frac {1}{1-X}}\right]=\operatorname {E} \left[\left({\frac {1}{1-X}}-\operatorname {E} \left[{\frac {1}{1-X}}\right]\right)^{2}\right]=\operatorname {var} \left[{\frac {X}{1-X}}\right]=} $
${\displaystyle \operatorname {E} \left[\left({\frac {X}{1-X}}-\operatorname {E} \left[{\frac {X}{1-X}}\right]\right)^{2}\right]={\frac {\alpha (\alpha +\beta -1)}{(\beta -2)(\beta -1)^{2}}}{\text{ if }}\beta >2} $
A: It suffices to find the standard deviation of $Y_{i}$. To this end first define a random variable $W\sim\text{Gamma}(p)$ ($p>0$) if $W$ has density
$$
f_W(w)=\frac{1}{\Gamma(p)}w^{p-1}e^{-w}\quad (w>0).
$$
It is easy to see that $EW^d=\frac{\Gamma(p+d)}{\Gamma(p)}$ by the definition of the gamma function. Now $X_i\stackrel{d}{=} X$ where
$$
X=\frac{Z_{1}}{Z_{1}+Z_{2}};\quad Z_{1}\perp Z_{2};\quad  Z_{1}\sim \text{Gamma}(6), Z_{2}\sim \text{Gamma}(4). 
$$
Then
$$
Y_{i}\stackrel{d}{=} \frac{Z_{1}/(Z_{1}+Z_{2)}}{Z_{2}/(Z_{1}+Z_{2)}}=\frac{Z_{1}}{Z_{2}}.
$$
Now
$$
\text{Var}(Y_{i})=EZ_{1}^{2}Z_{2}^{-2}-(EZ_{1}Z_{2}^{-1})^2.
$$
But then using the fact that $Z_{1}\perp Z_{2}$ we have that
$$
EZ_{1}^{2}Z_{2}^{-2}=EZ_{1}^{2}\cdot EZ_{2}^{-2}=\frac{7(6)}{3(2)}=7
$$
and
$$
EZ_{1}Z_{2}^{-1}=EZ_{1}\cdot EZ_{2}^{-1}=\frac{6}{3}=2.
$$
Thus 
$$
\text{Var}(Y_{i})=7-2^2=3\implies\sigma(Y_{i})=\sqrt{3}.
$$
The answer is (C).
