In a triangle ABC, if A + 2 B + C = 2 and the lengths of the side… — Solution | JEE Mathematics

MCQMediumJEE 2023Properties of Triangles

In a triangle $ABC$ , if $\cos A + 2 \cos B + \cos C = 2$ and the lengths of the sides opposite to the angles $A$ and $C$ are $3$ and $7$ respectively, then $\cos A - \cos C$ is equal to:

A
$\frac{3}{7}$
B
$\frac{9}{7}$
C
$\frac{10}{7}$
D
$\frac{5}{7}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: In triangle $ABC$ ,

\cos A + 2\cos B + \cos C = 2

and the sides opposite to $A$ and $C$ are $a = 3$ and $c = 7$ .

Find: $\cos A - \cos C$ .

From the provided solution, we use

\cos\left(\frac{A+C}{2}\right) = \sin\left(\frac{B}{2}\right)

and it concludes that after simplification,

\cos A - \cos C = \frac{10}{7}

Therefore, the correct option is C.

Common mistakes

Using the given relation $\cos A + 2\cos B + \cos C = 2$ without triangle angle identities is incorrect. In a triangle, angle-sum relations are essential. Use $A+B+C=\pi$ along with trigonometric identities.
Confusing side lengths with cosine values is wrong. The numbers $3$ and $7$ are the sides opposite $A$ and $C$ , not the values of $\cos A$ and $\cos C$ . Use them only through triangle side-angle relationships.
Applying the formula for $\cos A - \cos C$ with the wrong sign is a common error. Use the correct difference identity carefully and simplify step by step before substituting values.

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Step-by-step solution

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