Q66 / 150 · MHT CET 2023 (13 May Shift 1) · Mathematics

The value of $\begin{aligned} \cos \left(18^{\circ}-\mathrm{A}\right) \cdot \cos ( & \left.18^{\circ}+\mathrm{A}\right) \\ & -\cos \left(72^{\circ}-\mathrm{A}\right) \cos \left(72^{\circ}+\mathrm{A}\right) \text { is }\end{aligned}$

  • A
    $\cos 72^{\circ}$
  • Correct answer
    B
    $\sin 54^{\circ}$
  • C
    $\sin 18^{\circ}$
  • D
    $\cos 54^{\circ}$

Solution

$\begin{aligned} & \cos \left(18^{\circ}-\mathrm{A}\right) \cos \left(18^{\circ}+\mathrm{A}\right) \\ & \quad-\cos \left(72^{\circ}-\mathrm{A}\right) \cos \left(72^{\circ}+\mathrm{A}\right) \\ & =\cos \left(18^{\circ}-\mathrm{A}\right) \cos \left[90^{\circ}-\left(72^{\circ}-\mathrm{A}\right)\right] \\ & \quad-\cos \left(72^{\circ}-\mathrm{A}\right) \cos \left[90^{\circ}-\left(18^{\circ}-\mathrm{A}\right)\right] \\ & =\sin \left(72^{\circ}-\mathrm{A}\right) \cos \left(18^{\circ}-\mathrm{A}\right) \\ & \quad \quad-\cos \left(72^{\circ}-\mathrm{A}\right) \sin \left(18^{\circ}-\mathrm{A}\right) \\ & =\sin \left[\left(72^{\circ}-\mathrm{A}\right)-\left(18^{\circ}-\mathrm{A}\right)\right] \\ & =\sin 54^{\circ}\end{aligned}$

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

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