JEE Main 2024 — Trigonometric Equations Question with Solution

Given: $\cos 2x+a·\sin x=2a-7$

$\Rightarrow 1-2{\sin }^{2}x+a\times \sin x=2a-7$

$\Rightarrow a\times \sin x-2a=-8+2{\sin }^{2}x$

$\Rightarrow a\left(\sin x-2\right)=2\left({\sin }^{2}x-4\right)$

$\Rightarrow a(\sin x-2)=2(\sin x-2)(\sin x+2)$

$\Rightarrow \sin x=2 \left[\text{Not possible}\right], a=2(\sin x+2)$

We know that, $-1\le \sin x\le 1$

$\Rightarrow \mathrm{a}\in [2,6]$

$\Rightarrow p=2, q=6$

Now, $r=\tan 9^{°}+\cot 9^{°}-\tan {27}^{°}-\cot {27}^{°}$

$\Rightarrow r=\frac{{\cos }^2{9}^{°}+{\sin }^2{9}^{°}}{\sin 9^{°}·\cos 9^{°}}-\frac{{\cos }^2{27}^{°}+{\sin }^2{27}^{°}}{\sin {27}^{°}·\cos {27}^{°}}$

$\Rightarrow r=\frac{1}{\sin 9^{°}·\cos 9^{°}}-\frac{1}{\sin {27}^{°}·\cos {27}^{°}}$

$\Rightarrow r=\frac{2}{2\times \sin 9^{°}·\cos 9^{°}}-\frac{2}{2\times \sin {27}^{°}·\cos {27}^{°}}$

$\Rightarrow r=\frac{2}{\sin {18}^{°}}-\frac{2}{\sin {54}^{°}}$

$\Rightarrow r=2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}\right]$

$\Rightarrow r=8\left[\frac{\sqrt{5}+1-\sqrt{5}+1}{5-1}\right]$

$\Rightarrow r=8\left(\frac{2}{4}\right)$

$\Rightarrow r=4$

$\Rightarrow p·q·r=2\times 6\times 4$

$\Rightarrow p·q·r=48$