If A. B. C are the angles of a triangle, then $\frac{\sin A+\sin B+\sin C}{\sin ^2 \frac{A}{2}-\sin ^2…

If A. B. C are the angles of a triangle, then $\frac{\sin A+\sin B+\sin C}{\sin ^2 \frac{A}{2}-\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}-1}=$
  1. $-2 \tan \frac{B}{2}$
  2. $-2 \cot \frac{B}{2}$
  3. $2 \tan \frac{B}{2}$
  4. $2 \cot \frac{B}{2}$

Solution

$\begin{aligned} & \frac{\sin A+\sin B+\sin C}{\sin ^2 \frac{A}{2}-\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}-1} \\ = & \frac{2 \sin \frac{A+B}{2} \cdot \cos \frac{A-B}{2}+2 \sin \frac{C}{2} \cdot \cos \frac{C}{2}}{\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)-\cos ^2 \frac{C}{2}} \\ = & \frac{2 \sin \left(\frac{\pi}{2}-\frac{C}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)+2 \sin \frac{C}{2} \cos \frac{C}{2}}{\sin \left(\frac{\pi}{2}-\frac{C}{2}\right) \sin \left(\frac{A-B}{2}\right)-\cos ^2 \frac{C}{2}} \\ = & \frac{2 \cos \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)+\sin \frac{C}{2}\right]}{\cos \frac{C}{2}\left[\sin \left(\frac{A-B}{2}\right)-\cos \frac{C}{2}\right]} \\ = & \frac{2 \cos \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)+\cos \frac{A+B}{2}\right]}{\cos \frac{C}{2}\left[\sin \left(\frac{A-B}{2}\right)-\sin \left(\frac{A+B}{2}\right)\right]} \\ = & \frac{2 \cos \frac{C}{2} \cdot 2 \cos \frac{A}{2} \cdot \cos \frac{B}{2}}{-\cos \frac{C}{2} \cdot 2 \cdot \cos \frac{A}{2} \cdot \sin \frac{B}{2}}=-2 \cot \frac{B}{2} .\end{aligned}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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