$\operatorname{Cos}\left(36^{\circ}-\mathrm{A}\right) \cos \left(36^{\circ}+\mathrm{A}\right)+\cos…
$\operatorname{Cos}\left(36^{\circ}-\mathrm{A}\right) \cos \left(36^{\circ}+\mathrm{A}\right)+\cos \left(54^{\circ}+\mathrm{A}\right) \cos \left(54^{\circ}-\mathrm{A}\right)=$
- $\operatorname{Cos} 2 \mathrm{~A}$
- $\operatorname{Cos} \mathrm{A}$
- $\operatorname{Sin} 2 \mathrm{~A}$
- $\operatorname{Sin} \mathrm{A}$
Solution
$\begin{array}{l}
\cos \left(36^{\circ}-\mathrm{A}\right) \cos \left(36^{\circ}+\mathrm{A}\right)+\cos \left(54^{\circ}+\mathrm{A}\right) \cos \left(54^{\circ}-\mathrm{A}\right) \\
=\cos \left(36^{\circ}-\mathrm{A}\right) \cdot \cos \left(36^{\circ}+\mathrm{A}\right)+\cos \left[90^{\circ}-\left(36^{\circ}-\mathrm{A}\right)\right] \cdot \cos \left[90^{\circ}-\left(36^{\circ}+\mathrm{A}\right)\right]
\end{array}$
$=\cos \left(36^{\circ}-\mathrm{A}\right) \cdot \cos \left(36^{\circ}+\mathrm{A}\right)+\sin \left(36^{\circ}-\mathrm{A}\right) \cdot \sin \left(36^{\circ}+\mathrm{A}\right)$
$=\cos \left[\left(36^{\circ}-\mathrm{A}\right)-\left(36^{\circ}+\mathrm{A}\right)\right]=\cos \left[36^{\circ}-\mathrm{A}-36^{\circ}-\mathrm{A}\right]=\cos (-2 \mathrm{~A})=\cos 2 \mathrm{~A}$
Asked in: MHT CET 2020 (16 Oct Shift 2)
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