Question

In \[\Delta ABC,(b - c)\cot \dfrac{A}{2} + (c - a)\cot \dfrac{B}{2} + (a - b)\cot \dfrac{C}{2}\]
is equal to
A. \[0\]
B. \[1\]
C. \[ \pm 1\]
D. \[2\]

Answer 1

Hint:In this case, we have been given a triangle ABC and we are to find the given equation \[(b - c)\cot \dfrac{A}{2} + (c - a)\cot \dfrac{B}{2} + (a - b)\cot \dfrac{C}{2}\] is equal to what value. For that we have to first use the sine formula \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k\] to initiate problem followed by equating each term to k and solving accordingly to the given equation to obtain the desired solution.
FORMULA USED:
We can use sine formula to solve these types of triangle problems,
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k\]

Complete step by step solution:We have been provided in the question that,
\[(b - c)\cot \dfrac{A}{2} + (c - a)\cot \dfrac{B}{2} + (a - b)\cot \dfrac{C}{2}\]
We know that the sine formula is as below,
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k\]
Now, we have to equate each term to \[k\] we obtain
\[a = k\sin A\]
\[b = k\sin B\]
\[c = k\sin C\]
Now, according to the given equation, write as below\[ = {\rm{k}}\left[ {(\sin {\rm{B}} - \sin {\rm{C}})\cot \left( {\dfrac{{\rm{A}}}{2}} \right) + (\sin {\rm{C}} - \sin {\rm{A)}}\cot \left( {\dfrac{{\rm{B}}}{2}} \right) + (\sin {\rm{A}} - \sin {\rm{B)}}\cot \left( {\dfrac{{\rm{C}}}{2}} \right)} \right]\]
Now, we have to apply \[(\sin {\rm{C}} - \sin {\rm{D = 2sin}}\left\{ {\left( {\dfrac{{{\rm{C}} - {\rm{D}}}}{2}} \right)} \right\}\cos \left( {\dfrac{{{\rm{C}} - {\rm{D}}}}{2}} \right)\] we get
\[ \Rightarrow k\left[ {2\sin \left( {\dfrac{{B - C}}{2}} \right)\cos \left( {\dfrac{{B + C}}{2}} \right)\cot \left( {\dfrac{A}{2}} \right) + 2\sin \left( {\dfrac{{C - A}}{2}} \right)\cos \left( {\dfrac{{C + A}}{2}} \right)\cot \left( {\dfrac{B}{2}} \right) + 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)\cot \left( {\dfrac{C}{2}} \right)} \right]\]
We have been already known that the sum of angles in a triangle is \[{180^0}\]
So, we can write as,
\[A + B + C = \pi \]
Now, we have to equate each term to \[\pi \] we get
\[\dfrac{{A{\rm{ }}}}{2} = \dfrac{\pi }{2} - \dfrac{{\left( {B + C} \right)}}{2}\]
\[\dfrac{{B{\rm{ }}}}{2} = \dfrac{\pi }{2} - \dfrac{{\left( {C + A} \right)}}{2}\]
\[\dfrac{{{\rm{C }}}}{2} = \dfrac{\pi }{2} - \dfrac{{\left( {A + B} \right)}}{2}\]
Now, we have to substitute values we obtained previously, we get\[ \Rightarrow k\left[ {2\sin \left( {\dfrac{{B - C}}{2}} \right)\sin \left( {\dfrac{{C + B}}{2}} \right) + 2\sin \left( {\dfrac{{C - A}}{2}} \right)\sin \left( {\dfrac{{A + C}}{2}} \right) + 2\sin \left( {\dfrac{{A - B}}{2}} \right)\sin \left( {\dfrac{{B + A}}{2}} \right)} \right]\]
Now, let us apply the formula \[\cos C - \cos D{\rm{ }} = {\rm{ }}2\sin \dfrac{{C + D}}{2}\sin \dfrac{{D - C}}{2}\] we get
\[ = k\left[ {\cos C - \cos B + \cos A - \cos C + \cos B - \cos A} \right]\]
On solving the terms inside the parentheses, all the similar terms get cancelled, we get
\[ \Rightarrow k(0)\]
On further simplification, we get
\[ = 0\]
Therefore, in \[\Delta ABC,(b - c)\cot \dfrac{A}{2} + (c - a)\cot \dfrac{B}{2} + (a - b)\cot \dfrac{C}{2}\]
 is equal to \[0\]

So, Option ‘A’ is correct

Note:Students are likely to make mistake in these types of problems because it includes many formulas like \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k\] and while solving these types of problems one should be very familiar with trigonometry trivial identities to get the required solution without getting it wrong. As these problems includes more signs and trigonometry terms it should be solved correctly by applying correct formulas to get the correct solution.