In any triangle ABC, prove the following:

a(cos B cos C + cos A) = b(cos C cos A + cos B) = c(cos A cos B + cos C)

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

So by considering the LHS of the given equation, we get

a(cos B cos C + cos A)

Now substituting the values from sine rule, we get

⇒ = k sin A(cos B cos C + cos A)

⇒ = k (sin A cos B cos C + sin A cos A)

Now consider the second part from the given equation, we get

b(cos A cos C + cos B)

Now substituting the values from sine rule, we get

⇒ = k sin B(cos A cos C + cos B)

⇒ = k (sin B cos A cos C + sin B cos B)

Now consider the third part from the given equation, we get

c(cos A cos B + cos C)

Now substituting the values from sine rule, we get

⇒ = k sin C(cos A cos B + cos C)

⇒ = k (sin C cos A cos B + sin C cos C)

From equation (i), (ii), and (iii), we get

a(cos B cos C + cos A) = b(cos C cos A + cos B) = c(cos A cos B + cos C)

Hence proved