In a $\triangle A B C, \frac{a-b}{a+b}=$

In a $\triangle A B C, \frac{a-b}{a+b}=$
  1. $\cot \left(\frac{A-B}{2}\right) \cot \frac{C}{2}$
  2. $\tan \left(\frac{A+B}{2}\right) \tan \frac{C}{2}$
  3. $\tan \left(\frac{A-B}{2}\right) \tan \frac{C}{2}$
  4. $\tan \left(\frac{A+B+C}{2}\right)$

Solution

$ \begin{aligned} & \frac{a-b}{a+b}=\frac{k \sin A-k \sin B}{k \sin A+k \sin B} apply sine rule \\ & =\frac{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)} \\ & =\cot \left(\frac{A+B}{2}\right) \tan \left(\frac{A-B}{2}\right) \\ & =\cot \left(90^{\circ}-\frac{c}{2}\right) \tan \left(\frac{A-B}{2}\right) \\ & =\tan \frac{C}{2} \tan \left(\frac{A-B}{2}\right) \end{aligned} $

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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