Q66 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

In a $\triangle A B C,\left(b^2-c^2\right) \cot A+\left(c^2-a^2\right) \cot B=$

  • A
    0
  • Correct answer
    B
    $2 R^2[\sin 2 A-\sin 2 B]$
  • C
    $\left(b^2-a^2\right) \cot (A+B)$
  • D
    $2 R^2[\tan 2 A-\tan 2 B]$

Solution

In $\triangle A B C$, $\left(b^2-c^2\right) \cot A+\left(c^2-a^2\right) \cot B$ We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ $\because 4 R^2\left[\left(\sin ^2 B-\sin ^2 C\right) \cot A+\left(\sin ^2 C-\sin ^2 A\right) \cot B\right]$ $4 R^2[\sin (B+C) \sin (B-C)]$ $\left.\frac{\cos A}{\sin A}+\sin (C+A) \sin (C-A) \frac{\cos B}{\sin B}\right]$ $4 R^2\left[\frac{\sin A \sin (B-C) \cos A}{\sin A}+\frac{\sin B \sin (C-A) \cos B}{\sin B}\right]$ $4 R^2[-\sin (B-C) \cos (B+C)-\sin (C-A) \cos (C+A)]$ $2 R^2[-\sin 2 B+\sin 2 C+\sin 2 A-\sin 2 C]$ $2 R^2[\sin 2 A-\sin 2 B]$

Hence, the correct answer is option: 'B'.

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