Q63 / 160 · TS EAMCET 2006 · Mathematics

$\sin 120^{\circ} \cos 150^{\circ}-\cos 240^{\circ} \sin 330^{\circ}$ is equal to :

  • A
    1
  • Correct answer
    B
    -1
  • C
    $\frac{2}{3}$
  • D
    $-\left(\frac{\sqrt{3}+1}{4}\right)$

Solution

$\sin 120^{\circ} \cos 150^{\circ}-\cos 240^{\circ} \sin 330^{\circ}$ $=-\cos 30^{\circ} \sin 60^{\circ}-\cos 60^{\circ} \sin 30^{\circ}$ $=-\left(\sin 60^{\circ} \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}\right)$ $=-\sin \left(60^{\circ}+30^{\circ}\right)=-\sin 90^{\circ}$ $=-1$

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

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