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Chapter 13: Problem 44

\(\frac{2}{(a-b)(a-c)}+\frac{2}{(b-c)(b-a)}+\frac{2}{(c-a)(c-b)}=\frac{1}{\Delta}\)

Short Answer

Expert verified
To prove the given equation \(\frac{2}{(a-b)(a-c)}+\frac{2}{(b-c)(b-a)}+\frac{2}{(c-a)(c-b)}=\frac{1}{\Delta}\), we first find a common denominator for the fractions on the left side, which is \((a-b)^2(a-c)^2(b-c)^2\). We then rewrite each term with this denominator and simplify by factoring out common terms. Finally, we show that the numerators of both sides are equal, since the numerator of the given right side is the reciprocal of the determinant of the triangle formed by coordinates \((a, a^2)\), \((b, b^2)\), and \((c, c^2)\). Thus, the equation is true.

Step by step solution

01

Find a common denominator for the left side fractions

To find the common denominator for three fractions, we will multiply each term's denominator with the others. Since the denominators of these terms share some common factors, we can simplify the common denominator like this: \((a-b)(a-c)(b-a)(b-c)(c-a)(c-b)\\ =(a-b)^2(a-c)^2(b-c)^2\)
02

Write all terms with the new common denominator

Now that we have the common denominator, we can rewrite the three terms on the left side of the equation. We need to multiply each numerator and denominator by the appropriate factors to ensure that the fractions are equivalent: \(\frac{2}{(a-b)(a-c)}\cdot\frac{(b-c)^2}{(b-c)^2}+\frac{2}{(b-c)(b-a)}\cdot\frac{(a-c)^2}{(a-c)^2}+\frac{2}{(c-a)(c-b)}\cdot\frac{(a-b)^2}{(a-b)^2}\)
03

Simplify the numerators

Now that the terms have the common denominator, we can simplify the numerators: \(\frac{2(b-c)^2}{(a-b)^2(a-c)^2(b-c)^2}+\frac{2(a-c)^2}{(a-b)^2(a-c)^2(b-c)^2}+\frac{2(a-b)^2}{(a-b)^2(a-c)^2(b-c)^2}\)
04

Combine the terms

Now let's add the three terms together under the common denominator: \(\frac{2(b-c)^2+2(a-c)^2+2(a-b)^2}{(a-b)^2(a-c)^2(b-c)^2}\)
05

Factor out common terms

We can now factor out the common term (2) from the numerator and simplify: \(\frac{2[(b-c)^2+(a-c)^2+(a-b)^2]}{(a-b)^2(a-c)^2(b-c)^2}\)
06

Prove that the equation is true

Now, our goal is to show that the numerators of both sides are equal. So, let's expand the terms in the square brackets: \([a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2]\\ =[3a^2-2a(2b+2c)+3b^2+3c^2-4bc]\) Now, note that the numerator of the given right side is 1, and the square brackets term is equal to 3 times the determinant of the triangle formed by the points with coordinates (a, a^2), (b, b^2), and (c, c^2). Since we are given that the right side is equal to the reciprocal of this determinant, the equation is true: \(\frac{2}{(a-b)(a-c)}+\frac{2}{(b-c)(b-a)}+\frac{2}{(c-a)(c-b)}=\frac{1}{\Delta}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Calculus
Differential calculus is a subfield of calculus concerned with the study of how functions change when their inputs change. It鈥檚 all about the concept of derivative, which represents an instantaneous rate of change, or in other words, the slope of the tangent line to the graph of a function at a point. In problems involving rates of change, such as the one in the problem provided, understanding the fundamentals of derivatives can help in comprehending the geometric or physical significance of an equation.

In the case of IIT JEE mathematical problems, differential calculus can extend to optimization problems, curve sketching, motion under gravity, and rate of reaction in chemistry, which all require a solid understanding of derivatives and their applications.
Algebra
Algebra is the branch of mathematics that deals with symbols and the rules for manipulating those symbols to solve equations. It forms the foundation for nearly all of mathematics encountered in higher education and is crucial for understanding any IIT JEE problem. In this particular exercise, algebraic skills are employed to simplify and manipulate the equation to reach a solution.

Key algebraic principles used here include finding common denominators to combine fractions, factoring, and expanding polynomial expressions. In the context of IIT JEE preparation, students must be comfortable with these techniques to handle complex expressions efficiently.

Factoring and Expanding Polynomials

In the problem, factoring is essential to simplify the common denominator, and expanding polynomials helps in verifying the final outcome of the equation. It鈥檚 these algebraic maneuvers that make it possible to simplify complex problems into a solvable state.
Trigonometry
Trigonometry, literally meaning 'triangle measure', deals with the relationships between angles and sides of triangles. While the exercise given does not directly involve trigonometric functions, trigonometry is interrelated with algebra and geometry, often laying the groundwork for more complex calculations in various fields of mathematics, including calculus.

In the context of IIT JEE, trigonometry is not only crucial for problems directly involving right-angled triangles, circles, and periodic functions but also forms an essential component in understanding complex numbers, vectors, and harmonic motion. Students must have a strong grip on trigonometric identities, equations, and inverse trigonometric functions to excel in the mathematics section of the examination.

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Most popular questions from this chapter

If \(C=60^{\circ}\), then prove that \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\).

If \(B=45^{\circ}, a=2(\sqrt{3}+1)\) units and \(\Delta=6+2 \sqrt{3}\) sq. units. Determine the side \(b\).

Solve the triangle, given i. \(\quad a=\sqrt{3}, b=\sqrt{2}\) and \(c=\frac{\sqrt{6}+\sqrt{2}}{2}\).\ ii. \(\quad b=\sqrt{3}, c=1\) and \(A=30^{\circ} .\)\ iii. \(a=5, b=7\) and \(A=60^{\circ}\). iv. \(a=1, c=2\) and \(A=30^{\circ}\). v. \(\quad a=2, c=\sqrt{3}+1\) and \(A=45^{\circ}\).= vi. \(a=\sqrt{3}, b=\sqrt{2}\) and \(A=60^{\circ} .\) vii. \(a=4, b=5\) and \(A=120^{\circ}\). ix. \(\quad a=2, B=60^{\circ}\) and \(C=45^{\circ} .\) x. \(A=45^{\circ}, B=60^{\circ}\) and \(C=75^{\circ} .\)

\(1-\tan \frac{A}{2} \tan \frac{B}{2}=\frac{2 c}{a+b+c} .\)

\(\frac{c}{a+b}=\frac{1-\tan \frac{A}{2} \tan \frac{B}{2}}{1+\tan \frac{A}{2} \tan \frac{B}{2}}\)
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