bcosC +c cosb = a

Since abc is a triangle, then we know that:

b^2 = a^2+c^2-2ac cos(b)

==> cosb=-(b^2-a^2-c^2)/2ac

and c^2 = a^2+b^2 -2ab cos(c)

==> cos (c)= -(c^2-a^2-b^2)ab

Now we need to prove that b cos(c)+ C cos(b) = a

then,

-b(c^2-a^2-b^2)/2ab -C (b^2-a^2-c^2)/2ac

= -(c^2 -a^2-b^2)/2a - (b^2-a^2-c^2)/2a

= (-c^2+a^2+b^2-b^2+a^2+c^2)/2a

= 2a^2/2a = a

Let ABC be the triangle. Draw a perpendicular from A to BC to meet BC at D. Then,

BC = a = BD+DC.

But BD = AB cos B from the right angled triangleABD . And DC = AC cos C from the right angled triangle ACD. Therefore,

BC = a = BD+DC = c*cos B+b*CosC. So the given equality is true for any triangle and is therefore an identity.

We'll prove this with the help from the cosine theorem:

b^2=c^2+a^2-2ac*cosB, cos B = (c^2+a^2-b^2)/2ac

c^2=a^2+b^2-2ab*cosC, cos C = (a^2+b^2-c^2)/2ab

Now, we'll substitute cos B and cos C, by the found expressions, so that:

b*cosC+c*cosB=a

b*(a^2+b^2-c^2)/2ab + c*(c^2+a^2-b^2)/2ac = a

After reducing the terms, we'll obtain the irreducible quotients:

(a^2+b^2-c^2)/2a + (c^2+a^2-b^2)/2a = a

We'll multiply the term from the right side, by 2a:

a^2+b^2-c^2+c^2+a^2-b^2 = 2a^2

After reducing similar terms, we'll obtain:

a^2+a^2 = 2a^2

2a^2 = 2a^2

The last result tells us that the identity, b*cosC+c*cosB=a, is valid for any triangle.