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Question: Answered & Verified by Expert
In any $\triangle A B C, \frac{b-c \cos A}{c-b \cos A}$ is equal to
MathematicsProperties of TrianglesTS EAMCETTS EAMCET 2021 (06 Aug Shift 2)
Options:
  • A $\frac{\sin B}{\sin C}$
  • B $\frac{\cos C}{\cos B}$
  • C $\frac{\cos B}{\cos C}$
  • D $\frac{\sin C}{\sin B}$
Solution:
1450 Upvotes Verified Answer
The correct answer is: $\frac{\cos C}{\cos B}$
$\begin{aligned} \frac{b-c \cos A}{c-b \cos A} & =\frac{b-\frac{b^2+c^2-a^2}{2 b}}{c-\frac{b^2+c^2-a^2}{2 c}} \\ & =\left(\frac{b^2+a^2-c^2}{c^2+a^2-b^2}\right) \times \frac{c}{b} \\ & =\frac{b^2+a^2-c^2}{2 a b} \cdot \frac{2 a c}{c^2+a^2-b^2} \\ & =\frac{\cos C}{\cos B}\end{aligned}$

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