
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
162.9k+ 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.
Recently Updated Pages
JEE Advanced 2021 Physics Question Paper 2 with Solutions

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

JEE Advanced 2022 Chemistry Question Paper 2 with Solutions

JEE Advanced 2025 Revision Notes for Chemistry Energetics - Free PDF Download

JEE Advanced Marks vs Rank 2025 - Predict IIT Rank Based on Score

JEE Advanced 2022 Maths Question Paper 2 with Solutions

Trending doubts
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

Top IIT Colleges in India 2025

IIT Fees Structure 2025

IIT Roorkee Average Package 2025: Latest Placement Trends Updates

Other Pages
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations
