Q.In $\triangle ABC$ with the usual notations prove that $(a - b)^2 \cos^2\left(\dfrac{C}{2}\right) + (a + b)^2 \sin^2\left(\dfrac{C}{2}\right) = c^2$.

Writing $a+b+c=2s$ (so $s$ is the semi-perimeter), the half-angle formulae express the sine, cosine and tangent of half of any angle of a triangle purely in terms of its three sides: $\sin\frac A2=\sqrt{\frac{(s-b)(s-c)}{bc}}$, $\cos\frac A2=\sqrt{\frac{s(s-a)}{bc}}$, and $\tan\frac A2=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ (with the exactly analogous formulae for $B$ and $C$). They are derived by substituting the Cosine Rule's expression for $\cos A$ into the double-angle identities $1-\cos A=2\sin^2\frac A2$ and $1+\cos A=2\cos^2\frac A2$, then factoring the resulting difference or sum of squares in terms of $s$. Heron's Formula, $A(\triangle ABC)= …