TS EAMCET · Maths · Properties of Triangles
In a \(\triangle A B C,\left(b^2-c^2\right) \cot A+\left(c^2-a^2\right) \cot B=\)
- A 0
- B \(2 R^2[\sin 2 A-\sin 2 B]\)
- C \(\left(b^2-a^2\right) \cot (A+B)\)
- D \(2 R^2[\tan 2 A-\tan 2 B]\)
Answer & Solution
Correct Answer
(B) \(2 R^2[\sin 2 A-\sin 2 B]\)
Step-by-step Solution
Detailed explanation
In \(\triangle A B C\), \(\left(b^2-c^2\right) \cot A+\left(c^2-a^2\right) \cot B\) We know that, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R\) \(\because 4 R^2\left[\left(\sin ^2 B-\sin ^2 C\right) \cot A+\left(\sin ^2 C-\sin ^2 A\right) \cot B\right]\)…
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