Q62 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics

If $A+B+C=60^{\circ}$, then $\cos \left(30^{\circ}-A\right)+\cos \left(30^{\circ}-B\right)+\cos \left(30^{\circ}-C\right)$ $+\sin (A+B+C)=$

  • Correct answer
    A
    $4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
  • B
    $4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
  • C
    $4 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$
  • D
    $4 \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$

Solution

For $A+B+C=60^{\circ}$, then $\left(\cos 30^{\circ}-A\right)+\cos \left(30^{\circ}-B\right)+\cos \left(30^{\circ}-C\right)$ $+\sin (A+B+C)$ $=2 \cos \left(30^{\circ}-\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\cos \left(30^{\circ}-C\right)$ $+\cos 30^{\circ}$ $=2 \cos \frac{C}{2} \cos \frac{A-B}{2}+2 \cos \left(30^{\circ}-\frac{C}{2}\right) \cos \left(\frac{C}{2}\right)$ $=2 \cos \frac{C}{2}\left[\cos \frac{A-B}{2}+\cos \left(30^{\circ}-\frac{C}{2}\right)\right]$ $=2 \cos \frac{C}{2}$ $\left[2 \cos \left(\frac{A-B-C}{4}+15^{\circ}\right) \cos \left(\frac{A-B+C}{4}-15^{\circ}\right)\right]$ $=4 \cos \frac{C}{2} \cos \left(\frac{A+60^{\circ}-B-C}{4}\right) \cos \left(\frac{A+C-60^{\circ}-B}{4}\right)$ $=4 \cos \frac{C}{2} \cos \left(\frac{2 A}{4}\right) \cos \left(\frac{-2 B}{4}\right)=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$

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

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