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.   

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).
