In triangle ABC, \[\left( {b + c} \right)\cos A + \left( {c + a} \right)\cos B + \left( {a + b} \right)\cos C\] is equal to
A) 0
B) 1
C) $\left( {a + b + c} \right)$
D) $2\left( {a + b + c} \right)$

Hint: First open the brackets and multiply the terms. After that group the terms in such a way that the projection law of triangles can be applied on the terms. For $\left( {b\cos C + c\cos B} \right)$, substitute $a$, for $\left( {a\cos C + c\cos A} \right)$ substitute $b$ and, for $\left( {a\cos B + b\cos A} \right)$ substitute $c$. The output will be the desired result.

Formula used: Projection law or the formula of projection law states that the algebraic sum of the projection of any two sides is equal to the third side of the triangle. The formula is:
$a = b\cos C + c\cos B$
$b = a\cos C + c\cos A$
$c = a\cos B + b\cos A$
where, A, B, C are the three angles of the triangle and a, b, c is the corresponding opposite side of the angles.

Complete step-by-step answer:
Now,
\[\left( {b + c} \right)\cos A + \left( {c + a} \right)\cos B + \left( {a + b} \right)\cos C\]
Open the brackets and multiply the terms,
$b\cos A + c\cos A + c\cos B + a\cos B + a\cos C + b\cos C$
Group the terms,
$\left( {b\cos C + c\cos B} \right) + \left( {a\cos C + c\cos A} \right) + \left( {a\cos B + b\cos A} \right)$
Now, apply the projection law of the triangle,
$a + b + c$
Thus, \[\left( {b + c} \right)\cos A + \left( {c + a} \right)\cos B + \left( {a + b} \right)\cos C\] is equal to $a + b + c$.

Hence, option (C) is correct.

Note: The students might make mistakes by not applying the projection formula of triangles which will lead to failure in finding the value.
The solution of triangles is the most important trigonometric problem to solve the triangle or to find out the characteristics of the triangles like measurements of the sides, angles of triangles while having some known values. When we need to find out the angles and all the sides of the triangle using the following rules:
The law of sines:- The formula is $\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$.
The law of cosines:- The formula is
$\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
$\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}$
$\cos C = \dfrac{{{b^2} + {a^2} - {c^2}}}{{2ab}}$
The projection formula:- The formula is
$a = b\cos C + c\cos B$
$b = a\cos C + c\cos A$
$c = a\cos B + b\cos A$
The sum of all the angles of a triangle is always equal to 180 degrees or π radians. The formula is $A + B + C = 180^\circ $.