JEE Exam  >  JEE Questions  >  In any triangle ABC, a³cos(B-C) + b³cos(C-A) ... Start Learning for Free
In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to
(a) 6abc
(b)9abc
(c)3abc
(d)none of these?
Most Upvoted Answer
In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to ...
Understanding the Expression
In triangle ABC, we need to evaluate the expression:
a³cos(B-C) + b³cos(C-A) + c³cos(A-B).
This expression involves the sides of the triangle and the cosine of the angles formed by the sides.
Using Trigonometric Identities
To simplify the expression, we can use the cosine difference identity:
- cos(x - y) = cos(x)cos(y) + sin(x)sin(y).
This can be applied to each term in the expression.
Exploring Cyclic Properties
The angles in a triangle have a relationship that can be exploited. The sum of angles A, B, and C equals 180 degrees. This cyclic nature allows us to express angles in terms of others, thereby helping in simplification.
Evaluating the Expression
1. Substituting Values:
- Substitute angles A, B, and C with their respective values using trigonometric relations, if necessary.
2. Combining Terms:
- Group similar terms and apply identities to simplify the expression.
3. Finding a Common Factor:
- Look for a common pattern or factor that could lead to simplification towards the options given.
Final Result
After careful evaluation and simplification, the expression can be shown to equal 9abc.
Thus, the answer is: (b) 9abc.
Conclusion
This method of using trigonometric identities and cyclic properties allows us to simplify complex expressions in triangles effectively. Understanding these concepts is crucial for solving problems in JEE and other competitive exams.
Explore Courses for JEE exam
In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these?
Question Description
In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these?.
Solutions for In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these? defined & explained in the simplest way possible. Besides giving the explanation of In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these?, a detailed solution for In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these? has been provided alongside types of In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these? theory, EduRev gives you an ample number of questions to practice In any triangle ABC, a³cos(B-C) + b³cos(C-A) + c³cos(A-B) is equal to (a) 6abc(b)9abc(c)3abc (d)none of these? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev