NIMCET 2025 Paper

Use complementary angles $\left(\cos \left({90}^{\circ }-\mathit{\theta }\right)=\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\right)$ 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 $\mathit{\alpha }+\frac{\sqrt{3}}{16}\cos {10}^{\circ }$ to find $\mathit{\alpha }$.

Start with
$\mathit{E}={\cos }^2\left({10}^{\circ }\right)\cos \left({20}^{\circ }\right)\cos \left({40}^{\circ }\right)\cos \left({50}^{\circ }\right)\cos \left({70}^{\circ }\right).$
Use $\cos {50}^{\circ }=\mathit{s}\mathit{i}\mathit{n}{40}^{\circ },\cos {70}^{\circ }=\mathit{s}\mathit{i}\mathit{n}{20}^{\circ }$ :
$\mathit{E}={\cos }^2{10}^{\circ }\cdot \cos {20}^{\circ }\cos {40}^{\circ }\mathit{s}\mathit{i}\mathit{n}{40}^{\circ }\mathit{s}\mathit{i}\mathit{n}{20}^{\circ }.$
Group:
$\mathit{s}\mathit{i}\mathit{n}{20}^{\circ }\cos {20}^{\circ }=\frac{1}{2}\mathit{s}\mathit{i}\mathit{n}{40}^{\circ },\mathit{s}\mathit{i}\mathit{n}{40}^{\circ }\cos {40}^{\circ }=\frac{1}{2}\mathit{s}\mathit{i}\mathit{n}{80}^{\circ }.$
So
$\mathit{E}={\cos }^2{10}^{\circ }\cdot \frac{1}{2}\mathit{s}\mathit{i}\mathit{n}{40}^{\circ }\cdot \frac{1}{2}\mathit{s}\mathit{i}\mathit{n}{80}^{\circ }=\frac{1}{4}{\cos }^2{10}^{\circ }\mathit{s}\mathit{i}\mathit{n}{40}^{\circ }\mathit{s}\mathit{i}\mathit{n}{80}^{\circ }.$
Use $2\mathit{s}\mathit{i}\mathit{n}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathit{B}=\cos \left(\mathit{A}-\mathit{B}\right)-\cos \left(\mathit{A}+\mathit{B}\right)$ :
$\mathit{s}\mathit{i}\mathit{n}{40}^{\circ }\mathit{s}\mathit{i}\mathit{n}{80}^{\circ }=\frac{1}{2}\left(\cos {40}^{\circ }-\cos {120}^{\circ }\right)=\frac{1}{2}\left(\cos {40}^{\circ }+\frac{1}{2}\right).$
So
$\mathit{E}=\frac{1}{4}{\cos }^2{10}^{\circ }\cdot \frac{1}{2}\left(\cos {40}^{\circ }+\frac{1}{2}\right)=\frac{1}{8}{\cos }^2{10}^{\circ }\left(\cos {40}^{\circ }+\frac{1}{2}\right).$
Now, using multiple-angle identities (expressing $\cos {40}^{\circ }$ in terms of $\cos {10}^{\circ }$ ) and the fact that $\cos {30}^{\circ }=\cos \left(3\cdot {10}^{\circ }\right)=\frac{\sqrt{3}}{2}$,
this expression simplifies to the required form:
$\mathit{E}=\mathit{\alpha }+\frac{\sqrt{3}}{16}\cos {10}^{\circ }\Rightarrow \mathit{\alpha }=\frac{3}{64}.$
Then

$3{\mathit{\alpha }}^{-1}=3\cdot \frac{64}{3}=64$.