Question

In a triangle \[ABC\], find the value of \[\left( {{b^2} - {c^2}} \right)\cot A + \left( {{c^2} - {a^2}} \right)\cot B + \left( {{a^2} - {b^2}} \right)\cot C\].
A. 0
B. \[{a^2} + {b^2} + {c^2}\]
C. \[2\left( {{a^2} + {b^2} + {c^2}} \right)\]
D. \[\dfrac{1}{{2abc}}\]

Answer 1

Hint: First, we will apply the basic trigonometric ratio \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\] . Then, substitute the values in the equation by using the laws of sines and cosines. Solve the equation and get the required answer.

Formula used:
Trigonometric ratio: \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
In a triangle \[ABC\],
Law of sines: \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\]
Law of cosines:
\[\cos A = \dfrac{{{b^2} + c{}^2 - {a^2}}}{{2bc}}\]
\[\cos B = \dfrac{{{c^2} + a{}^2 - {b^2}}}{{2ac}}\]
\[\cos C = \dfrac{{{a^2} + b{}^2 - {c^2}}}{{2ab}}\]

Complete step by step solution:
The triangle \[ABC\] is given.

Let’s solve the given equation.
Apply the basic ratio of trigonometry \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\].
\[\left( {{b^2} - {c^2}} \right)\cot A + \left( {{c^2} - {a^2}} \right)\cot B + \left( {{a^2} - {b^2}} \right)\cot C = \left( {{b^2} - {c^2}} \right)\dfrac{{\cos A}}{{\sin A}} + \left( {{c^2} - {a^2}} \right)\dfrac{{\cos B}}{{\sin B}} + \left( {{a^2} - {b^2}} \right)\dfrac{{\cos C}}{{\sin C}}\]
Substitute the values of sine and cosine angles by using the laws of sines and cosines.
We get,
\[\left( {{b^2} - {c^2}} \right)\cot A + \left( {{c^2} - {a^2}} \right)\cot B + \left( {{a^2} - {b^2}} \right)\cot C = \left( {{b^2} - {c^2}} \right)\dfrac{{\dfrac{{{b^2} + c{}^2 - {a^2}}}{{2bc}}}}{{ka}} + \left( {{c^2} - {a^2}} \right)\dfrac{{\dfrac{{{c^2} + a{}^2 - {b^2}}}{{2ac}}}}{{kb}} + \left( {{a^2} - {b^2}} \right)\dfrac{{\dfrac{{{a^2} + b{}^2 - {c^2}}}{{2ab}}}}{{kc}}\]
\[ \Rightarrow \left( {{b^2} - {c^2}} \right)\cot A + \left( {{c^2} - {a^2}} \right)\cot B + \left( {{a^2} - {b^2}} \right)\cot C = \left( {{b^2} - {c^2}} \right)\dfrac{{{b^2} + c{}^2 - {a^2}}}{{2kabc}} + \left( {{c^2} - {a^2}} \right)\dfrac{{{c^2} + a{}^2 - {b^2}}}{{2kabc}} + \left( {{a^2} - {b^2}} \right)\dfrac{{{a^2} + b{}^2 - {c^2}}}{{2kabc}}\]
Factor out the common terms from the right-hand side.
\[\left( {{b^2} - {c^2}} \right)\cot A + \left( {{c^2} - {a^2}} \right)\cot B + \left( {{a^2} - {b^2}} \right)\cot C = \dfrac{1}{{2kabc}}\left[ {\left( {{b^2} - {c^2}} \right)\left( {{b^2} + c{}^2 - {a^2}} \right) + \left( {{c^2} - {a^2}} \right)\left( {{c^2} + a{}^2 - {b^2}} \right) + \left( {{a^2} - {b^2}} \right)\left( {{a^2} + b{}^2 - {c^2}} \right)} \right]\]
Solve the right-hand side of the above equation.
\[\left( {{b^2} - {c^2}} \right)\cot A + \left( {{c^2} - {a^2}} \right)\cot B + \left( {{a^2} - {b^2}} \right)\cot C = \dfrac{1}{{2kabc}}\left[ {\left( {{b^4} - {c^4} - {a^2}{b^2} + {a^2}c{}^2} \right) + \left( {{c^4} - {a^4} - {b^2}{c^2} + {a^2}b{}^2} \right) + \left( {{a^4} - {b^4} - {a^2}{c^2} + {b^2}c{}^2} \right)} \right]\]
\[ \Rightarrow \left( {{b^2} - {c^2}} \right)\cot A + \left( {{c^2} - {a^2}} \right)\cot B + \left( {{a^2} - {b^2}} \right)\cot C = \dfrac{1}{{2kabc}}\left[ 0 \right]\]
\[ \Rightarrow \left( {{b^2} - {c^2}} \right)\cot A + \left( {{c^2} - {a^2}} \right)\cot B + \left( {{a^2} - {b^2}} \right)\cot C = 0\]
Hence the correct option is A.

Note: The law of sine or the sine law states that the ratio of the side length of a triangle to the sine of the opposite angle, is the same for all three sides.
The law of cosines is used to find an unknown side of a triangle given the value of two sides and their included angle or to find an unknown angle given three sides of a triangle.