TS EAMCET · Maths · Properties of Triangles
In a \(\triangle A B C, \Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right)\) is equal to
- A a
- B b
- C c
- D 0
Answer & Solution
Correct Answer
(D) 0
Step-by-step Solution
Detailed explanation
\begin{aligned} & \Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) \\ & \therefore(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) \\ & =(b+c) \cdot \frac{(b-c)}{(b+c)} \cot \frac{A}{2} \tan \frac{A}{2} \\ & =b-c \\ & \therefore \quad \Sigma(b+c) \tan…
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