JEE 2023 — Mathematics PYQ

$4{\cos }^2\theta -1=4(1-{\sin }^2\theta )-1$  

$=3-4{\sin }^2\theta$  

$=\frac{3\sin \theta -4{\sin }^3\theta }{\sin \theta }$  

$=\frac{\sin 3\theta }{\sin \theta }$  

So,  

$36(4{\cos }^2{9}^{\circ }-1)(4{\cos }^{2}27^{\circ }-1)(4{\cos }^{2}81^{\circ }-1)(4{\cos }^{2}243^{\circ }-1)$  

$=36\left[\frac{\sin 27^{\circ }}{\sin 9^{\circ }}\times \frac{\sin 81^{\circ }}{\sin 27^{\circ }}\times \frac{\sin 243^{\circ }}{\sin 81^{\circ }}\times \frac{\sin 729^{\circ }}{\sin 243^{\circ }}\right]$  

$=36\left[\frac{\sin 729^{\circ }}{\sin 9^{\circ }}\right]=36\times 1=36$