Question

In \[\Delta ABC\], \[\left( {a\cot A + b\cot B + c\cot C} \right)\] (where a, b, c are the sides of a triangle) is equals to
(A) \[2R + r\]
(B) \[R + 2r\]
(C) \[R + r\]
(D) \[2\left( {R + r} \right)\]

Answer 1

Hint: In this question, we have to choose the required specific form of the given particulars.
First we will apply sine formula in the given particular and then using the trigonometric formulas we will get the required result.

Formula used: According to the law, \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R\]
Here a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the opposite angles (see the figure to the right), while R is the radius of the triangle's circumcircle.
\[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
\[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \]
\[\sin A\sin B = \dfrac{1}{2}(\cos (A - B) + \cos (A + B))\]
\[\left( {\cos A + \cos B} \right) = \left( {2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}} \right)\]
\[\cos C = 1 - {\sin ^2}\dfrac{C}{2}\]

Complete step-by-step answer:
We need to find out in the \[\Delta ABC\], \[\left( {a\cot A + b\cot B + c\cot C} \right)\] (where a,b,c are the sides of a triangle).
using sine formula, \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R\],
We can equate this formula and we can write it as,
$\Rightarrow$\[\dfrac{a}{{\sin A}} = 2R\]
Taking cross multiplication we get,
$\Rightarrow$\[a = 2R\sin A\]
Similarly we can write it as
$\Rightarrow$\[\dfrac{b}{{\sin B}} = 2R\]
Taking cross multiplication we get,
$\Rightarrow$\[b = 2R\sin B\]
Also, we can write it as.
$\Rightarrow$\[\dfrac{c}{{\sin C}} = 2R\]
Taking cross multiplication we get,
$\Rightarrow$\[c = 2R\sin C\]
Putting the value in the given equation,
$\Rightarrow$\[\left( {a\cot A + b\cot B + c\cot C} \right)\]\[ = 2R\sin A\cot A + 2R\sin B\cot B + 2R\sin C\cot C\]
We use from trigonometric formula, \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\], using it we get,
\[ \Rightarrow 2R\sin A\dfrac{{\cos A}}{{\sin A}} + 2R\sin B\dfrac{{\cos B}}{{\sin B}} + 2R\sin C\dfrac{{\cos C}}{{\sin C}}\]
Cancelling the same term and we get,
\[ \Rightarrow 2R\left( {\cos A + \cos B + \cos C} \right)\]
Using the formula, and we get,
\[ \Rightarrow 2R\left( {2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} + 1 - {{\sin }^2}\dfrac{C}{2}} \right)\]
Since A,B,C are the angles of a triangle ,so \[A + B + C = \pi \]
\[ \Rightarrow 2R\left\{ {1 + 2\cos \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right)\cos \dfrac{{A - B}}{2} - {{\sin }^2}\dfrac{C}{2}} \right\}\]
Now we use the formula that,\[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \]
\[ \Rightarrow 2R\left\{ {1 + 2\sin \left( {\dfrac{C}{2}} \right)\cos \dfrac{{A - B}}{2} - {{\sin }^2}\dfrac{C}{2}} \right\}\]
Taking common we get,
\[ \Rightarrow 2R\left\{ {1 + 2\sin \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} - \sin \dfrac{C}{2}} \right)} \right\}\]
Since A, B, C are the angles of a triangle, so \[A + B + C = \pi \], using it we get,
\[ \Rightarrow 2R\left\{ {1 + 2\sin \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} - \sin \left( {\dfrac{\pi }{2} - \dfrac{{A + B}}{2}} \right)} \right)} \right\}\]
Now we use the formula that, \[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \]
\[ \Rightarrow 2R\left\{ {1 + 2\sin \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} - \cos \dfrac{{A + B}}{2}} \right)} \right\}\]
Using the formula, \[\sin A\sin B = \dfrac{1}{2}(\cos (A - B) + \cos (A + B))\]we get,
\[ \Rightarrow 2R\left\{ {1 + 2\sin \dfrac{C}{2} \times 2\sin \dfrac{A}{2}\sin \dfrac{B}{2}} \right\}\]
Now multiply the term and we get,
\[ \Rightarrow 2R\left\{ {1 + 4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}} \right\}\]
We know that, \[4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2} = \dfrac{r}{R}\]
So, we can write it as,
 \[2R\left( {1 + 4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}} \right) = 2R\left( {1 + \dfrac{r}{R}} \right)\]
Taking LCM on the bracket term and we get,
\[ = 2R\dfrac{{R + r}}{R}\]
On cancelling the term,
\[ = 2\left( {R + r} \right)\]

Thus (D) is the correct option.

Note: In trigonometry, the law of sines, sine law, sine formula, or sine rule is an equation relating the lengths of the sides of a triangle (any shape) to the sines of its angles.
 According to the law, \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R\]
Where a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the opposite angles, while R is the radius of the triangle's circumcircle.