Q256 / 270 · NDA 2012 (Phase 1) · Mathematics

What is the value of $\sin 420^{\circ} \cdot \cos 390^{\circ}+\cos \left(-300^{\circ}\right) \cdot \sin \left(-330^{\circ}\right) ?$

  • A
    0
  • Correct answer
    B
    1
  • C
    2
  • D
    -1

Solution

$\begin{aligned} & \sin \left(360^{\circ}+60^{\circ}\right) \cdot \cos \left(360^{\circ}+30^{\circ}\right)+\cos \left(360^{\circ}-60^{\circ}\right)\{- \\ & \left.\sin \left(360^{\circ}-30^{\circ}\right)\right\} \\ & =\sin 60^{\circ} \cdot \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ} \\ & =\sin \left(60^{\circ}+30^{\circ}\right) \\ & =\sin 90^{\circ}=1\end{aligned}$

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

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