MHT CET · Maths · Trigonometric Equations
The value of
\(\cos \left(18^{\circ}-\mathrm{A}\right) \cdot \cos (\left.18^{\circ}+\mathrm{A}\right) \ -\cos \left(72^{\circ}-\mathrm{A}\right)\cos\) \(\left(72^{\circ}+\mathrm{A}\right) \text { is }\)
- A \(\cos 72^{\circ}\)
- B \(\sin 54^{\circ}\)
- C \(\sin 18^{\circ}\)
- D \(\cos 54^{\circ}\)
Answer & Solution
Correct Answer
(B) \(\sin 54^{\circ}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \cos \left(18^{\circ}-\mathrm{A}\right) \cos \left(18^{\circ}+\mathrm{A}\right) \\ & \quad-\cos \left(72^{\circ}-\mathrm{A}\right) \cos \left(72^{\circ}+\mathrm{A}\right) \\ & =\cos \left(18^{\circ}-\mathrm{A}\right) \cos \left[90^{\circ}-\left(72^{\circ}-\mathrm{A}\right)\right] \\ & \quad-\cos \left(72^{\circ}-\mathrm{A}\right) \cos \left[90^{\circ}-\left(18^{\circ}-\mathrm{A}\right)\right] \\ & =\sin \left(72^{\circ}-\mathrm{A}\right) \cos \left(18^{\circ}-\mathrm{A}\right) \\ & \quad \quad-\cos \left(72^{\circ}-\mathrm{A}\right) \sin \left(18^{\circ}-\mathrm{A}\right) \\ & =\sin \left[\left(72^{\circ}-\mathrm{A}\right)-\left(18^{\circ}-\mathrm{A}\right)\right] \\ & =\sin 54^{\circ}\end{aligned}\)
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