MCQ
If $\tan \frac{ B - C }{2}=x \cot \frac{A}{2}$, then $x=$
  • A
    $\frac{c-a}{c+a}$
  • B
    $\frac{a-b}{a+b}$
  • $\frac{b-c}{b+c}$
  • D
    $\frac{ b - c }{ a }$

Answer

Correct option: C.
$\frac{b-c}{b+c}$
(C) By Napier's analogy, we have
$\tan \frac{ B - C }{2}=\frac{ b - c }{ b + c } \cot \frac{ A }{2} \Rightarrow x=\frac{ b - c }{ b + c }$

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