SOLUTION: In triangle ABC, the value of acotA+bcotB+ccotC is? a)R+r b)(R+r)/R c)2(R+r) d)3(R+r) Also explain how?

1.  a*cot(A) + b*cot(B) + c*cot(C) =  = .    (1)


     Next,   = 2R,    = 2R  and   = 2R,  where R is the radius of the circumscribed circle around the triangle,

     according to the Sine Law theorem (see the lessons Law of sines  and  Law of sines - the Geometric Proof in this site).



    Therefore, the line (1) can be continued in this way

    a*cot(A) + b*cot(B) + c*cot(C) = 2R*(sin(A) + sin(B) + sin(C)).     (2)


    It is the first idea, and it allows us to reduce the problem to calculation of  sin(A) + sin(B) + sin(C).



2.  The second idea is  THIS:


        For any triangle with angles  A, B and C

        sin(A) + sin(B) + sin(C) = ,                              (3)  

        where r is the radius of the inscribed circle, while R is the radius of the circumscribed circle about the triangle.


    Deriving formula (3) requires some technique, but it is known proof, which you can find at this reference

    https://math.stackexchange.com/questions/734395/how-to-prove-that-fracrr1-cos-a-cos-b-cos-c



3.  Finally,  a*cot(A) + b*cot(B) + c*cot(C) = 2R*(r/R + 1)}}} = 2*(R+r).