Prove that in any triangle $ABC$,$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$,where $a, b, c$ are the lengths of the sides opposite to the vertices $A, B, C$ respectively.

(N/A) Consider a triangle $ABC$ with sides $a, b, c$ opposite to vertices $A, B, C$. Draw a perpendicular $BD$ from vertex $B$ to side $AC$. Let $D$ be the point on $AC$ such that $BD \perp AC$.
In the right-angled triangle $ABD$,we have:
$AD = c \cos A$
$BD = c \sin A$
Since $AC = b$,the length $CD = AC - AD = b - c \cos A$.
Now,in the right-angled triangle $BDC$,by the Pythagorean theorem:
$BC^{2} = BD^{2} + CD^{2}$
$a^{2} = (c \sin A)^{2} + (b - c \cos A)^{2}$
Expanding the terms:
$a^{2} = c^{2} \sin^{2} A + b^{2} + c^{2} \cos^{2} A - 2bc \cos A$
Using the identity $\sin^{2} A + \cos^{2} A = 1$:
$a^{2} = b^{2} + c^{2}(\sin^{2} A + \cos^{2} A) - 2bc \cos A$
$a^{2} = b^{2} + c^{2} - 2bc \cos A$
Rearranging the terms to solve for $\cos A$:
$2bc \cos A = b^{2} + c^{2} - a^{2}$
$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$