ABC is a triangle, then find \[{a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\cos }^2}C - {{\cos }^2}A} \right) + {c^2}\left( {{{\cos }^2}A - {{\cos }^2}B} \right)\].
A. 0
B. 1
C. \[{a^2} + {b^2} + {c^2}\]
D. \[2\left( {{a^2} + {b^2} + {c^2}} \right)\]
Answer
261.3k+ views
Hint: First we will convert all \[\cos \] function into \[\sin \] function using trigonometry identity. Then using the sine law, we will find the value of \[\sin A\], \[\sin B\] and \[\sin C\]. After that we will substitute the value of \[\sin A\], \[\sin B\] and \[\sin C\] in the given expression.
Formula used:
Sine law
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]
Trigonometry identity
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Complete step by step solution:
Given expression is \[{a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\cos }^2}C - {{\cos }^2}A} \right) + {c^2}\left( {{{\cos }^2}A - {{\cos }^2}B} \right)\]
Now applying the trigonometry identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[ = {a^2}\left( {1 - {{\sin }^2}B - 1 + {{\sin }^2}C} \right) + {b^2}\left( {1 - {{\sin }^2}C - 1 + {{\sin }^2}A} \right) + {c^2}\left( {1 - {{\sin }^2}A - 1 + {{\sin }^2}B} \right)\]
\[ = {a^2}\left( {{{\sin }^2}C - {{\sin }^2}B} \right) + {b^2}\left( {{{\sin }^2}A - {{\sin }^2}C} \right) + {c^2}\left( {{{\sin }^2}B - {{\sin }^2}A} \right)\] …..(i)
We know that,
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\left( {{\rm{say}}} \right)\]
\[\sin A = ak\], \[\sin B = bk\], and \[\sin C = ck\]
Substitute \[\sin A = ak\], \[\sin B = bk\], and \[\sin C = ck\] in the expression (i)
\[ = {a^2}\left( {{c^2}{k^2} - {b^2}{k^2}} \right) + {b^2}\left( {{a^2}{k^2} - {c^2}{k^2}} \right) + {c^2}\left( {{b^2}{k^2} - {a^2}{k^2}} \right)\]
Simplify the above expression
\[ = {a^2}{c^2}{k^2} - {a^2}{b^2}{k^2} + {b^2}{a^2}{k^2} - {b^2}{c^2}{k^2} + {c^2}{b^2}{k^2} - {c^2}{a^2}{k^2}\]
Cancel out the opposite term
\[ = 0\]
Hence option A is the correct option.
Note: Students often make a common mistake to solve the given question. They apply cosine law to solve the given question. But the given question should be solved by using trigonometry identity and the sine law.
Formula used:
Sine law
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]
Trigonometry identity
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Complete step by step solution:
Given expression is \[{a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\cos }^2}C - {{\cos }^2}A} \right) + {c^2}\left( {{{\cos }^2}A - {{\cos }^2}B} \right)\]
Now applying the trigonometry identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[ = {a^2}\left( {1 - {{\sin }^2}B - 1 + {{\sin }^2}C} \right) + {b^2}\left( {1 - {{\sin }^2}C - 1 + {{\sin }^2}A} \right) + {c^2}\left( {1 - {{\sin }^2}A - 1 + {{\sin }^2}B} \right)\]
\[ = {a^2}\left( {{{\sin }^2}C - {{\sin }^2}B} \right) + {b^2}\left( {{{\sin }^2}A - {{\sin }^2}C} \right) + {c^2}\left( {{{\sin }^2}B - {{\sin }^2}A} \right)\] …..(i)
We know that,
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\left( {{\rm{say}}} \right)\]
\[\sin A = ak\], \[\sin B = bk\], and \[\sin C = ck\]
Substitute \[\sin A = ak\], \[\sin B = bk\], and \[\sin C = ck\] in the expression (i)
\[ = {a^2}\left( {{c^2}{k^2} - {b^2}{k^2}} \right) + {b^2}\left( {{a^2}{k^2} - {c^2}{k^2}} \right) + {c^2}\left( {{b^2}{k^2} - {a^2}{k^2}} \right)\]
Simplify the above expression
\[ = {a^2}{c^2}{k^2} - {a^2}{b^2}{k^2} + {b^2}{a^2}{k^2} - {b^2}{c^2}{k^2} + {c^2}{b^2}{k^2} - {c^2}{a^2}{k^2}\]
Cancel out the opposite term
\[ = 0\]
Hence option A is the correct option.
Note: Students often make a common mistake to solve the given question. They apply cosine law to solve the given question. But the given question should be solved by using trigonometry identity and the sine law.
Recently Updated Pages
JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

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

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Electromagnetic Waves and Their Importance

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

