NIMCET 2025 Paper

Use complementary angles $(\cos (90^{\circ} - \theta) = \sin \; \theta)$ to convert some cosines to sines, then use product-to-sum formulas to simplify the product. Finally, express everything in terms of $\cos 10^{\circ}$ and compare with the given form $\alpha + \frac{\sqrt{3}}{16} \cos 10^{\circ}$ to find $\alpha$. Start with$E = \cos^2 (10^{\circ}) \cos (20^{\circ}) \cos (40^{\circ}) \cos (50^{\circ}) \cos (70^{\circ}) .$Use $\cos 50^{\circ} = \sin \; 40^{\circ}, \cos 70^{\circ} = \sin \; 20^{\circ}$ :$E = \cos^2 10^{\circ} \cdot \cos 20^{\circ} \cos 40^{\circ} \sin \; 40^{\circ} \sin \; 20^{\circ} .$Group:$\sin \; 20^{\circ} \cos 20^{\circ} = \frac{1}{2} \sin \; 40^{\circ}, \;\; \sin \; 40^{\circ} \cos 40^{\circ} = \frac{1}{2} \sin \; 80^{\circ} .$So$E = \cos^2 10^{\circ} \cdot \frac{1}{2} \sin \; 40^{\circ} \cdot \frac{1}{2} \sin \; 80^{\circ} = \frac{1}{4} \cos^2 10^{\circ} \sin \; 40^{\circ} \sin \; 80^{\circ} .$Use $2 \sin \; A \sin \; B = \cos (A - B) - \cos (A + B)$ :$\sin \; 40^{\circ} \sin \; 80^{\circ} = \frac{1}{2} (\cos 40^{\circ} - \cos 120^{\circ}) = \frac{1}{2} \left(\cos 40^{\circ} + \frac{1}{2}\right) .$So$E = \frac{1}{4} \cos^2 10^{\circ} \cdot \frac{1}{2} \left(\cos 40^{\circ} + \frac{1}{2}\right) = \frac{1}{8} \cos^2 10^{\circ} \left(\cos 40^{\circ} + \frac{1}{2}\right) .$Now, using multiple-angle identities (expressing $\cos 40^{\circ}$ in terms of $\cos 10^{\circ}$ ) and the fact that $\cos 30^{\circ} = \cos (3 \cdot 10^{\circ}) = \frac{\sqrt{3}}{2}$,this expression simplifies to the required form:$E = \alpha + \frac{\sqrt{3}}{16} \cos 10^{\circ} \;\; \Rightarrow \;\; \alpha = \frac{3}{64} .$Then $3 \alpha^{-1} = 3 \cdot \frac{64}{3} = 64$.