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QJEEM24Ma8e3d: JEE Main 2024 Mathematics - Class 11 JEEM24Ma8e3d Solution: Let the set of all such that the equation has a solution be and …

Let the set of all a∈Ra \in \mathbf{R} such that the equation cos⁡2x+asin⁡x=2a−7\cos 2 x+a \sin x=2 a-7 has a solution be [p,q][p, q] and r=tan⁡9∘−tan⁡27∘−1cot⁡63∘+tan⁡81∘r=\tan 9^{\circ}-\tan 27^{\circ}-\frac{1}{\cot 63^{\circ}}+\tan 81^{\circ} , then pqr is equal to ____________.

Step-by-Step Verified Solution

amp;cos⁡2x+a⋅sin⁡x=2a−7amp;a(sin⁡x−2)=2(sin⁡x−2)(sin⁡x+2)amp;sin⁡x=2,a=2(sin⁡x+2)amp;⇒a∈[2,6]amp;p=2q=6amp;r=tan⁡9∘+cot⁡9∘−tan⁡27−cot⁡27amp;r=1sin⁡9⋅cos⁡9−1sin⁡27⋅cos⁡27amp;=2[45−1−45+1]amp;r=4amp;p⋅q⋅r=2×6×4=48\begin{aligned}& \cos 2 x+a \cdot \sin x=2 a-7 \\& a(\sin x-2)=2(\sin x-2)(\sin x+2) \\& \sin x=2, a=2(\sin x+2) \\& \Rightarrow a \in[2,6] \\& p=2 \quad q=6 \\& r=\tan 9^{\circ}+\cot 9^{\circ}-\tan 27-\cot 27 \\& r=\frac{1}{\sin 9 \cdot \cos 9}-\frac{1}{\sin 27 \cdot \cos 27} \\& =2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}\right] \\& r=4 \\& p \cdot q \cdot r=2 \times 6 \times 4=48\end{aligned}

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