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JEE Main 2024
Trigonometry
Trigonometric Ratio and Identites
Medium

Question

Let the set of all aRa \in \mathbf{R} such that the equation cos2x+asinx=2a7\cos 2 x+a \sin x=2 a-7 has a solution be [p,q][p, q] and r=tan9tan271cot63+tan81r=\tan 9^{\circ}-\tan 27^{\circ}-\frac{1}{\cot 63^{\circ}}+\tan 81^{\circ}, then pqr is equal to ____________.

Answer: 2

Solution

cos2x+asinx=2a7a(sinx2)=2(sinx2)(sinx+2)sinx=2,a=2(sinx+2)a[2,6]p=2q=6r=tan9+cot9tan27cot27r=1sin9cos91sin27cos27=2[45145+1]r=4pqr=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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