With the usual notations, in a ∆ A B C , if a = 2 ,   b = 6 and c = 3 + 1 , then sin 2 C - sin 2…

We have, $a=2, b=\sqrt{6}$ and $c=\sqrt{3}+1$

Using cosine rule, we get,

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

$=\frac{6-2\sqrt{3}}{4\sqrt{6}}$

$=\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)$

And $\cos A=\frac{c^2+b^2-a^2}{2cb}$

$=\frac{4+2\sqrt{3}+6-4}{2\left(\sqrt{3}+1\right)\sqrt{6}}=\frac{1}{\sqrt{2}}$

Now,

${\sin }^{2}C-{\sin }^{2}A=\left(1-{\cos }^{2}C\right)-\left(1-{\cos }^{2}A\right)$

$={\cos }^{2}A-{\cos }^{2}C$

$=\frac{1}{2}-{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}^2$

$=\frac{1}{2}-\left(\frac{4-2\sqrt{3}}{8}\right)$

$=\frac{\sqrt{3}}{4}$.