Question

If the circumradius and inradius of a triangle be 10 and 3 respectively, then the value of a cot A + b cot B + c cot C is equal to 25.
(a) True
(b) False

Answer 1

Hint:First of all, draw the diagram to visualize the question. Now, write a = 2R sin A, b = 2R sin B, c = 2R sin C and also, $$\cot \theta ={\displaystyle \frac{\cos \theta }{\sin \theta }}$$ in the given expression. Now, proceed with solving the expression and finally substitute $$4\sin {\displaystyle \frac{A}{2}}\sin {\displaystyle \frac{B}{2}}\sin {\displaystyle \frac{C}{2}}={\displaystyle \frac{r}{R}}$$ to get the desired value.

Complete step-by-step answer:
We are given that the circumradius and inradius of a triangle be 10 and 3 respectively, then we have to verify if a cot A + b cot B + c cot C is equal to 25 or not. Let us see what inradius and circumradius are by the diagram.

In the above figure, OB is the circumradius (R) of $$\mathrm{\Delta }ABC$$ and OD is the inradius (r) of $$\mathrm{\Delta }ABC$$. Also, a, b and c are the sides of $$\mathrm{\Delta }ABC$$. Here, we are given that R = OB = 10. Also, R = OB = OC = OA = 10 as all the radii are of the same length.
We are also given that r = OD = 3. Now, let us consider the expression given in the question.
E = a cot A + b cot B + c cot C
We know that $$\cot \theta ={\displaystyle \frac{\cos \theta }{\sin \theta }}$$. So by using this in the above equation, we get,
$$E=a{\displaystyle \frac{\cos A}{\sin A}}+b{\displaystyle \frac{\cos B}{\sin B}}+c{\displaystyle \frac{\cos C}{\sin C}}$$
Now, we know that, a = 2R sin A, b = 2R sin B, c = 2R sin C. By substituting them in the above equation, we get,
$$E=2R\sin A{\displaystyle \frac{\cos A}{\sin A}}+2R\sin B{\displaystyle \frac{\cos B}{\sin B}}+2R\sin C{\displaystyle \frac{\cos C}{\sin C}}$$
By cancelling the like terms from the above equation, we get,
$$E=2R\cos A+2R\cos B+2R\cos C$$
$$E=2R(\cos A+\cos B+\cos C)$$
Now, we know that,
$$\cos x+\cos y=2\cos \left({\displaystyle \frac{x+y}{2}}\right)\cos \left({\displaystyle \frac{x-y}{2}}\right)$$
Also, $$\cos x=1-2{\sin }^2{\displaystyle \frac{x}{2}}$$. By using this in the above equation, we get,
$$E=2R[2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)+1-2{\sin }^2{\displaystyle \frac{C}{2}}]$$
We know that for any triangle,
$$A+B+C=\pi$$
$$A+B=\pi -C$$
By using this, we get,
$$E=2R[2\cos \left({\displaystyle \frac{\pi -C}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)+1-2{\sin }^2{\displaystyle \frac{C}{2}}]$$
We know that,
$$\cos ({\displaystyle \frac{\pi }{2}}-\theta )=\sin \theta$$
So, by using this, we get,
$$E=2R[2\sin {\displaystyle \frac{C}{2}}\cos \left({\displaystyle \frac{A-B}{2}}\right)+1-2{\sin }^2{\displaystyle \frac{C}{2}}]$$
We can write the above equation as,
$$E=2R[1+2\sin {\displaystyle \frac{C}{2}}(\cos \left({\displaystyle \frac{A-B}{2}}\right)-\sin {\displaystyle \frac{C}{2}})]$$
By replacing C by $$\pi -(A+B)$$ in the second bracket, we get,
$$E=2R[1+2\sin {\displaystyle \frac{C}{2}}(\cos \left({\displaystyle \frac{A-B}{2}}\right)-\sin ({\displaystyle \frac{\pi }{2}}-\left({\displaystyle \frac{A+B}{2}}\right)))]$$
We know that $$\sin ({\displaystyle \frac{\pi }{2}}-\theta )=\cos \theta$$. By using this, we get,
$$E=2R[1+2\sin {\displaystyle \frac{C}{2}}(\cos \left({\displaystyle \frac{A-B}{2}}\right)-\cos \left({\displaystyle \frac{A+B}{2}}\right))]$$
We know that,
$$\cos x-\cos y=-2\sin \left({\displaystyle \frac{x+y}{2}}\right)\sin \left({\displaystyle \frac{x-y}{2}}\right)$$
By using this, we get,
$$E=2R[1+2\sin {\displaystyle \frac{C}{2}}[-2\sin \left({\displaystyle \frac{\left({\displaystyle \frac{A-B}{2}}\right)+\left({\displaystyle \frac{A+B}{2}}\right)}{2}}\right)\sin \left({\displaystyle \frac{\left({\displaystyle \frac{A-B}{2}}\right)-\left({\displaystyle \frac{A+B}{2}}\right)}{2}}\right)]]$$
$$E=2R[1+2\sin {\displaystyle \frac{C}{2}}(-2\sin \left({\displaystyle \frac{A}{2}}\right)\sin \left({\displaystyle \frac{-B}{2}}\right))]$$
We know that $$\sin (-\theta )=-\sin \theta$$. By using this, we get,
$$E=2R[1+4\sin {\displaystyle \frac{C}{2}}.\sin {\displaystyle \frac{A}{2}}.\sin {\displaystyle \frac{B}{2}}]$$
Now, we know that for any triangle ABC, $${\displaystyle \frac{\text{inradius}}{\text{circumradius}}}$$=$4\sin {\displaystyle \frac{A}{2}}\sin {\displaystyle \frac{B}{2}}\sin {\displaystyle \frac{C}{2}}$
Or, $${\displaystyle \frac{r}{R}}=4\sin {\displaystyle \frac{A}{2}}\sin {\displaystyle \frac{B}{2}}\sin {\displaystyle \frac{C}{2}}$$
By using this we get,
$$E=2R(1+{\displaystyle \frac{r}{R}})$$
$$E=2R+2r$$
By substituting the values of R = 10 and r = 3. We get,
$$E=2\left(10\right)+2\left(3\right)$$
$$E=20+6$$
$$E=26$$
Hence, the value of a cot A + b cos B + c cot C = 26.
So, given the value of the expression is false.

Note: In these types of questions, students can directly remember that for a triangle ABC, (cos A + cos B + cos C) is equal to $$1+{\displaystyle \frac{r}{R}}$$. This is a very useful result in the solution of the triangle and would save a lot of steps. In this question, students often make mistakes while adding angles. So, this must be avoided.