Question 1089204
.
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?
~~~~~~~~~~~~~~~~~~~


It requires two ideas.


<pre>
1.  a*cot(A) + b*cot(B) + c*cot(C) = {{{a*(cos(A)/sin(A)) + b*(cos(b)/sin(b)) + c*(cos(C)/sin(C))}}} = {{{(a/sin(A))*cos(A) + (b/sin(B))*cos(B) + (c/sin(C))*cos(C))}}}.    (1)


     Next,  {{{a/sin(A)}}} = 2R,   {{{b/sin(B)}}} = 2R  and  {{{c/sin(C)}}} = 2R,  where R is the radius of the circumscribed circle around the triangle,

     according to the Sine Law theorem (see the lessons <A HREF=https://www.algebra.com/algebra/homework/Triangles/Law-of-sines.lesson>Law of sines</A>  and  <A HREF=https://www.algebra.com/algebra/homework/Triangles/Law-of-sines-the-Geometric-Proof.lesson>Law of sines - the Geometric Proof</A> 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  <U>THIS</U>:


        For any triangle with angles  A, B and C

        sin(A) + sin(B) + sin(C) = {{{r/R + 1}}},                              (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).
</pre>

Solved.