MHT CET · Maths · Properties of Triangles
In a triangle ABC with usual notations if, \(\tan \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=x \cot \frac{\mathrm{~A}}{2}\), then \(x=\)
- A \(\frac{c-a}{c+a}\)
- B \(\frac{a-b}{a+b}\)
- C \(\frac{b-c}{b+c}\)
- D \(\frac{a+b}{a-b}\)
Answer & Solution
Correct Answer
(C) \(\frac{b-c}{b+c}\)
Step-by-step Solution
Detailed explanation
\( \tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot \frac{A}{2} \) \( x \cot \frac{A}{2} = \frac{b-c}{b+c} \cot \frac{A}{2} \)
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