Q63 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

If $A+B+C=2 S$, then $\sin (S-A) \cos (S-B)-\sin (S-C) \cos S=$

  • A
    $\cos A \sin B \sin C$
  • B
    $\sin A \cos B \cos C$
  • Correct answer
    C
    $\cos A \sin B$
  • D
    $\sin A \cos B$

Solution

Since, $A+B+C=2 S$ $\quad ....\mathrm{(i)}$ $\begin{aligned} & \begin{array}{l}\frac{1}{2}[2 \sin (S-A) \cdot \cos (S-B)+2 \sin (S-C) \cdot \cos S]\end{array} \\ & \begin{array}{l}=\frac{1}{2}[\sin (S-A+S-B)+\sin (S-A-S+B) \\ \quad+\sin (S-C+S)+\sin (S-C-S)]\end{array} \\ & =\frac{1}{2}[\sin (2 S-A-B)+\sin (B-A)+\sin (2 S-C)-\sin C]\end{aligned}$ $\begin{aligned} & =\frac{1}{2}[\sin C+\sin (B-A)+\sin (A+B)-\sin C] \\ & =\frac{1}{2} \cdot 2 \sin \frac{B-A+A+B}{2} \cdot \cos \left(\frac{B-A-A-B}{2}\right) \\ & =\sin B \cdot \cos (-A)=\sin B \cos A\end{aligned}$

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

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