JEE Main 2023 — Mathematics Trigonometry

We need to find the value of

$$36(4{\mathrm{cos}}^{2}{9}^{\circ }-1)(4{\mathrm{cos}}^{2}{27}^{\circ }-1)(4{\mathrm{cos}}^{2}{81}^{\circ }-1)(4{\mathrm{cos}}^{2}{243}^{\circ }-1)$$36(4cos29∘−1)(4cos227∘−1)(4cos281∘−1)(4cos2243∘−1)

$$=36[(4{\mathrm{cos}}^{2}{9}^{\circ }-1)(4{\mathrm{sin}}^{2}{9}^{\circ }-1)(4{\mathrm{cos}}^{2}{27}^{\circ }-1)(4{\mathrm{sin}}^{2}{27}^{\circ }-1)]$$=36[(4cos29∘−1)(4sin29∘−1)(4cos227∘−1)(4sin227∘−1)]

$$=36[(4{(2\mathrm{cos}{9}^{\circ }\mathrm{sin}9^{\circ})}^{2}-4{\mathrm{cos}}^{2}9^{\circ}-4{\mathrm{sin}}^{2}9^{\circ}+1)(4{(2\mathrm{sin}{27}^{\circ }\mathrm{cos}9^{\circ})}^{2}-4{\mathrm{cos}}^{2}27^{\circ}-4{\mathrm{sin}}^{2}27^{\circ}+1)]$$=36[(4(2cos9∘sin9∘)2−4cos29∘−4sin29∘+1)(4(2sin27∘cos9∘)2−4cos227∘−4sin227∘+1)]

$$=36[(4{\mathrm{sin}}^{2}{18}^{\circ }-4+1)(4{\mathrm{sin}}^{2}{54}^{\circ }-4+1)]$$=36[(4sin218∘−4+1)(4sin254∘−4+1)]

$$=36[(4{\mathrm{sin}}^{2}{18}^{\circ }-3)(4{\mathrm{sin}}^{2}{54}^{\circ }-3)]$$=36[(4sin218∘−3)(4sin254∘−3)]

$$=36[{\frac{{(\sqrt{5}-1)}^{2}}{4}-3}{\frac{{(\sqrt{5}+1)}^{2}}{4}-3}]$$=36[4(5​−1)2​−34(5​+1)2​−3]

$$=36[{\frac{(6-2\sqrt{5}-12)}{4}\frac{(6+2\sqrt{5}-12)}{4}}]$$=36[4(6−25​−12)​4(6+25​−12)​]

$$=-36[{\frac{{(2\sqrt{5})}^{2}-36}{16}}]=36$$=−36[16(25​)2−36​]=36

Hence this is the correct option.