In a ABC ,(b²-c²) A +(c²-a²) B =

Correct Answer is : 2R2[sin⁡2A−sin⁡2B]2R²[ 2A- 2B]2R2[sin2A−sin2B]

Correct Answer is : 2R2[sin⁡2A−sin⁡2B]2R²[ 2A- 2B]2R2[sin2A−sin2B]

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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 26 of 160

Marks: +1, -0

In a

\triangle ABC ,(b^{2}-c^{2}) \cot A +(c^{2}-a^{2}) \cot B =

Solution:

\triangle A B C

(b^{2}-c^{2}) \cot A+(c^{2}-a^{2}) \cot B

We know that,

\;\frac{a}{\sin A}=\;\frac{b}{\sin B}=\;\frac{c}{\sin C}=2 R

\because 4 R^{2}[(\sin^{2} B-\sin^{2} C) \cot A+(\sin^{2} C-\sin^{2} A) \cot B]

4 R^{2}[\sin (B+C) \sin (B-C)]

\;\frac{\cos A}{\sin A}+\sin (C+A) \sin (C-A) \;\frac{\cos B}{\sin B}]

4 R^{2}\left[\;\frac{\sin A \sin (B-C) \cos A}{\sin A}+\;\frac{\sin B \sin (C-A) \cos B}{\sin B}\right]

4 R^{2}[-\sin (B-C) \cos (B+C)-\sin (C-A) \cos (C+A)]

2 R^{2}[-\sin 2 B+\sin 2 C+\sin 2 A-\sin 2 C]

2 R^{2}[\sin 2 A-\sin 2 B]

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