Combined mean for $n_!$ and $n_2$ observations are $ \bar {X} = \large\frac{n_1\bar{x}_1+n_2 \bar {x_2}}{n_1+n_2}$

to calculate standard deviation

$\sum x_1^2 = n_1 ( \sigma_1^2 +\bar {x}_1^2)$

$\sum x_2^2 = n_2 ( \sigma_2^2 +\bar {x}_2^2)$

Combined $SD= \sqrt { \large\frac{n_1 (s_1 ^2 + \bar{x}_1)+n_2 (s_2^2+\bar {x}_2^2)}{n_1+n_2} - \bigg( \large\frac{n_1 \bar {x_1}+n_2 \bar {x_2}}{n_1+n_2}\bigg)^2}$

$\sigma = \sqrt {\large\frac{n_1s_1^2 +n_2s_2^2}{n_1+n_2}+\frac{n_1\bar{x}_1^2+n_2x_2^2)(n_1n_2) -(n_1 \bar {x} _1 +n_2 \bar{x}_2)^2}{(n_1+n_2}}$

$\quad= \sqrt {\large\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2}+\frac{(n_1\bar{x_1}^2 +n_2 \bar {x_2}^2)(n_1+n_2)- (n_1 \bar {x_1} +n_2 \bar{x_2})^2}{(n_1+n_2)^2}}$

Expanding the equation

$\qquad= \sqrt{\large\frac{n_1s_1^2 +n_2 s_2^2}{n_1+n_2} +\frac{n_1^2 x_1^2 +n_1 \bar{x_2}^2 +n_2^2x_2^2-(n_1^2 x_1^2+n_2^2 x_2^2 +2 n_1n_2 \bar {x_1} \bar {x_2})}{(n_1+n_2)^2}}$

$\qquad= \sqrt {\large\frac{n_1 (s_1)^2 +n_2 (s_2)^2}{n_1+n_2} +\frac{n_1n_2(\bar {x_1} -\bar {x_2} )^2}{(n_1+n_2)^2}}$

Hence proved