JEE Main 2024 (Online) 27th January Morning ShiftQuestion 1
Let the set of all $a\in\mathbf{R}$ such that the equation $\cos2x+a\sin x=2a-7$ has a solution be $[p,q]$ and $r=\tan9^{\circ}-\tan27^{\circ}-\frac{1}{\cot63^{\circ}}+\tan81^{\circ}$, then $pqr$ is equal to $\_\_\_\_\_\_$ .
Trigonometric Ratio & Identities
Answer$48$
Solution
Concept. The equation factorises in $s=\sin x$, giving the range of $a$. For $r$, pair $\tan\theta$ with $\cot\theta$. Formulas used. (1) $\cos2x=1-2\sin^2x$(2) $\tan\theta+\cot\theta=\dfrac{2}{\sin2\theta}$, $\sin54^\circ-\sin18^\circ=2\cos36^\circ\sin18^\circ$Step 1 - Range of $a$. By (1), $1-2s^2+as=2a-7$, i.e. $2(2-s)(2+s)=a(2-s)$. Since $s\ne2$: $a=2s+4$ with $s\in[-1,1]$, so $a\in[2,6]$: $p=2$, $q=6$. Step 2 - $r$. $\frac{1}{\cot63^\circ}=\tan63^\circ=\cot27^\circ$ and $\tan81^\circ=\cot9^\circ$. By (2), $$r=(\tan9^\circ+\cot9^\circ)-(\tan27^\circ+\cot27^\circ)=\frac{2}{\sin18^\circ}-\frac{2}{\sin54^\circ}.$$ $$r=\frac{2(\sin54^\circ-\sin18^\circ)}{\sin18^\circ\sin54^\circ}=\frac{4\cos36^\circ\sin18^\circ}{\sin18^\circ\cos36^\circ}=4.$$ Step 3 - Value. $pqr=2\cdot6\cdot4=48$. Answer: 48