JEE Main 2023 — Trigonometric Ratios & Identities Question with Solution

We need to find the value of

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

$=36\left[\left(4 {\cos }^2{9}^{\circ }-1\right) \left(4 {\sin }^2{9}^{\circ }-1\right) \left(4 {\cos }^2{27}^{\circ }-1\right)\left(4 {\sin }^2{27}^{\circ }-1\right)\right]$

$=36\left[\left(4{\left(2\cos 9^{\circ }\sin 9°\right)}^2-4{\cos }^{2}9°-4{\sin }^{2}9°+1\right) \left(4{\left(2\sin {27}^{\circ }\cos 9°\right)}^2-4{\cos }^{2}27°-4{\sin }^{2}27°+1\right)\right]$

$=36\left[\left(4 {\sin }^2{18}^{\circ }-4 +1\right) \left(4 {\sin }^2{54}^{\circ }-4+1\right)\right]$

$=36\left[\left(4 {\sin }^2{18}^{\circ }-3\right) \left(4 {\sin }^2{54}^{\circ }-3\right)\right]$
$=36\left[\left\{\frac{{\left(\sqrt{5}-1\right)}^2}{4}-3\right\} \left\{\frac{{\left(\sqrt{5}+1\right)}^2}{4}-3\right\}\right]$
$=36\left[\left\{\frac{\left(6-2\sqrt{5}-12\right)}{4}\frac{\left(6+2\sqrt{5}-12\right)}{4}\right\}\right]$
$=-36\left[\left\{\frac{{\left(2\sqrt{5}\right)}^2-36}{16}\right\}\right]=36$

Hence this is the correct option.