AP EAMCET · Maths · Properties of Triangles
In a \(\triangle A B C, \frac{a-b}{a+b}=\)
- A \(\cot \left(\frac{A-B}{2}\right) \cot \frac{C}{2}\)
- B \(\tan \left(\frac{A+B}{2}\right) \tan \frac{C}{2}\)
- C \(\tan \left(\frac{A-B}{2}\right) \tan \frac{C}{2}\)
- D \(\tan \left(\frac{A+B+C}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\tan \left(\frac{A-B}{2}\right) \tan \frac{C}{2}\)
Step-by-step Solution
Detailed explanation
\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)…
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