The value of $\begin{aligned} & \cos \left(18^{\circ}-\mathrm{A} ight) \cos \left(18^{\circ}+\mathrm{A} ight) \ & -\cos \left(72^{\circ}-\mathrm{A} ight) \cos \left(72^{\circ}+\mathrm{A} ight) ext { is equal to } \end{aligned}$

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The value of $\begin{aligned} & \cos \left(18^{\circ}-\mathrm{A}ight) \cos \left(18^{\circ}+\mathrm{A}ight) \ & -\cos \left(72^{\circ}-\mathrm{A}ight) \cos \left(72^{\circ}+\mathrm{A}ight) 	ext { is equal to } \end{aligned}$

Aicharya · Accepted Answer

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

The value of $\begin{aligned} & \cos \left(18^{\circ}-\mathrm{A}\right) \cos…

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