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


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