If $\alpha$ and $\beta$ be the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of
$\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}$, then:
Correct Option: , 4
Using Binomial expansion
$(x+a)^{n}+(x-a)^{n}=2\left(T_{1}+T_{3}+T_{5}+T_{7} \cdots\right)$
$\therefore\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}=2\left(T_{1}+T_{3}+T_{5}+T_{7}\right)$
$2\left[{ }^{6} C_{0} x^{5}+{ }^{6} C_{2} x^{4}\left(x^{2}-1\right)+{ }^{6} C_{4} x^{2}\left(x^{2}-1\right)^{2}\right.$
$\left.+{ }^{6} C_{6}\left(x^{2}-1\right)^{3}\right]$
$=2\left[x^{6}+15\left(x^{6}-x^{4}\right)+15 x^{2}\left(x^{4}-2 x^{2}+1\right)\right.$
$\left.+\left(-1+3 x^{2}-3 x^{4}+x^{6}\right)\right]$
$=2\left(32 x^{6}-48 x^{4}+18 x^{2}-1\right)$
$\alpha=-96$ and $\beta=36$
$\therefore \quad \alpha-\beta=-132$