$$\sqrt{1+2t\sigma\cos\theta+\frac12t^2\sigma^2(3+\cos(2\theta))}=1+t\sigma\cos\theta+\frac12t^2\sigma^2+O(t^3)$$
What is the omitted step between these two equations? The parameter $t$ lies between $0$ and $1$ and it is said that approximation by using Taylor series is used.
Hint. One may use the Taylor series expansion, as $u \to 0$, $$ \sqrt{1+u}=1+\frac u2-\frac{u^2}{8}+O(u^3) $$ applying it to $u=2t\sigma \cos \theta+\dfrac 12t^2\sigma^2 (3+ \cos 2\theta) $ as $t \to 0$, observing that $$ \frac u2-\frac{u^2}{8}=t\sigma \cos \theta+\dfrac 14t^2\sigma^2 (3+ \cos 2\theta)-\frac{4t^2\sigma^2 \cos^2 \theta}{8}+O(t^3). $$