
In a triangle \[ABC\] , if \[{b^2} + {c^2} = 3{a^2}\] . Then what is the value of \[\cot B + \cot C - \cot A\]?
A. 1
B. \[\dfrac{{ab}}{{4\Delta }}\]
C. 0
D. \[\dfrac{{ac}}{{4\Delta }}\]
Answer
162k+ views
Hint: First, simplify the given equation by using the basic trigonometric ratio \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]. Then apply the laws of sines and cosines and solve the equation to get the required answer.
Formula used:
In a triangle \[ABC\]
Law of sines: \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\]
Laws 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:
Given:
In a triangle \[ABC\], \[{b^2} + {c^2} = 3{a^2}\].
Where \[a, b, c\] are the lengths of the sides of the triangle \[ABC\].
Let’s simplify the given trigonometric equation.
Apply the basic trigonometric ratio \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\].
\[\cot B + \cot C - \cot A = \dfrac{{\cos B}}{{\sin B}} + \dfrac{{\cos C}}{{\sin C}} - \dfrac{{\cos A}}{{\sin A}}\]
Apply the laws of sines and cosines.
\[\cot B + \cot C - \cot A = \dfrac{{\dfrac{{{c^2} + a{}^2 - {b^2}}}{{2ac}}}}{{kb}} + \dfrac{{\dfrac{{{a^2} + b{}^2 - {c^2}}}{{2ab}}}}{{kc}} - \dfrac{{\dfrac{{{b^2} + c{}^2 - {a^2}}}{{2bc}}}}{{ka}}\]
\[ \Rightarrow \cot B + \cot C - \cot A = \dfrac{{{c^2} + a{}^2 - {b^2}}}{{2kabc}} + \dfrac{{{a^2} + b{}^2 - {c^2}}}{{2kabc}} - \dfrac{{{b^2} + c{}^2 - {a^2}}}{{2kabc}}\]
\[ \Rightarrow \cot B + \cot C - \cot A = \dfrac{{3{a^2} - \left( {c{}^2 + {b^2}} \right)}}{{2kabc}}\]
Substitute the value from the given equation.
\[\cot B + \cot C - \cot A = \dfrac{{3{a^2} - 3{a^2}}}{{2kabc}}\]
\[ \Rightarrow \cot B + \cot C - \cot A = \dfrac{0}{{2kabc}}\]
\[ \Rightarrow \cot B + \cot C - \cot A = 0\]
Hence the correct option is C.
Note: Students often make mistakes while substituting the values in the sine ratio. They substitute \[\sin A = \dfrac{{a\sin B}}{b}\] or \[\sin A = \dfrac{{a\sin C}}{c}\] and other equivalent ratos. For this reason, they are unable to get the answer.
Formula used:
In a triangle \[ABC\]
Law of sines: \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\]
Laws 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:
Given:
In a triangle \[ABC\], \[{b^2} + {c^2} = 3{a^2}\].
Where \[a, b, c\] are the lengths of the sides of the triangle \[ABC\].
Let’s simplify the given trigonometric equation.
Apply the basic trigonometric ratio \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\].
\[\cot B + \cot C - \cot A = \dfrac{{\cos B}}{{\sin B}} + \dfrac{{\cos C}}{{\sin C}} - \dfrac{{\cos A}}{{\sin A}}\]
Apply the laws of sines and cosines.
\[\cot B + \cot C - \cot A = \dfrac{{\dfrac{{{c^2} + a{}^2 - {b^2}}}{{2ac}}}}{{kb}} + \dfrac{{\dfrac{{{a^2} + b{}^2 - {c^2}}}{{2ab}}}}{{kc}} - \dfrac{{\dfrac{{{b^2} + c{}^2 - {a^2}}}{{2bc}}}}{{ka}}\]
\[ \Rightarrow \cot B + \cot C - \cot A = \dfrac{{{c^2} + a{}^2 - {b^2}}}{{2kabc}} + \dfrac{{{a^2} + b{}^2 - {c^2}}}{{2kabc}} - \dfrac{{{b^2} + c{}^2 - {a^2}}}{{2kabc}}\]
\[ \Rightarrow \cot B + \cot C - \cot A = \dfrac{{3{a^2} - \left( {c{}^2 + {b^2}} \right)}}{{2kabc}}\]
Substitute the value from the given equation.
\[\cot B + \cot C - \cot A = \dfrac{{3{a^2} - 3{a^2}}}{{2kabc}}\]
\[ \Rightarrow \cot B + \cot C - \cot A = \dfrac{0}{{2kabc}}\]
\[ \Rightarrow \cot B + \cot C - \cot A = 0\]
Hence the correct option is C.
Note: Students often make mistakes while substituting the values in the sine ratio. They substitute \[\sin A = \dfrac{{a\sin B}}{b}\] or \[\sin A = \dfrac{{a\sin C}}{c}\] and other equivalent ratos. For this reason, they are unable to get the answer.
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