NIMCET 2025 — Mathematics PYQ

Let the Left-Hand Side (LHS) of the given equation be $E$.

$$E={\cos }^2(10^{\circ })\cos (20^{\circ })\cos (40^{\circ })\cos (50^{\circ })\cos (70^{\circ })$$

We can rewrite ${\cos }^2(10^{\circ })$ as $\cos (10^{\circ })\cdot \cos (10^{\circ })$ and rearrange the terms:

$$E=\cos (10^{\circ })\cdot \left[\cos (10^{\circ })\cos (50^{\circ })\cos (70^{\circ })\right]\cdot \left[\cos (20^{\circ })\cos (40^{\circ })\right]$$

Step 1: Simplify the first group using a standard product identity

We know the trigonometric identity:

$$\cos (\theta )\cos (60^{\circ }-\theta )\cos (60^{\circ }+\theta )=\frac{1}{4}\cos (3\theta )$$

If we put $\theta =10^{\circ }$:

$$\cos (10^{\circ })\cos (50^{\circ })\cos (70^{\circ })=\frac{1}{4}\cos (3\times 10^{\circ })=\frac{1}{4}\cos (30^{\circ })$$

Since $\cos (30^{\circ })=\frac{\sqrt{3}}{2}$:

$$\cos (10^{\circ })\cos (50^{\circ })\cos (70^{\circ })=\frac{1}{4}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{8}$$

Step 2: Substitute this value back into the expression

Now substitute this result back into our equation for $E$:

$$E=\cos (10^{\circ })\cdot \left(\frac{\sqrt{3}}{8}\right)\cdot \left[\cos (20^{\circ })\cos (40^{\circ })\right]$$

$$E=\frac{\sqrt{3}}{8}\cos (10^{\circ })\cos (20^{\circ })\cos (40^{\circ })$$

Step 3: Simplify the product of cosines

To simplify $\cos (10^{\circ })\cos (20^{\circ })\cos (40^{\circ })$, we can multiply and divide by $2\sin (10^{\circ })$:

$$\cos (10^{\circ })\cos (20^{\circ })\cos (40^{\circ })=\frac{2\sin (10^{\circ })\cos (10^{\circ })\cos (20^{\circ })\cos (40^{\circ })}{2\sin (10^{\circ })}$$

Using the double-angle formula $2\sin (\theta )\cos (\theta )=\sin (2\theta )$:

• $2\sin (10^{\circ })\cos (10^{\circ })=\sin (20^{\circ })$

The expression becomes:

$$=\frac{\sin (20^{\circ })\cos (20^{\circ })\cos (40^{\circ })}{2\sin (10^{\circ })}$$

Multiply and divide by $2$ again:

$$=\frac{2\sin (20^{\circ })\cos (20^{\circ })\cos (40^{\circ })}{4\sin (10^{\circ })}$$

$$=\frac{\sin (40^{\circ })\cos (40^{\circ })}{4\sin (10^{\circ })}$$

Multiply and divide by $2$ one last time:

$$=\frac{2\sin (40^{\circ })\cos (40^{\circ })}{8\sin (10^{\circ })}$$

$$=\frac{\sin (80^{\circ })}{8\sin (10^{\circ })}$$

Since $\sin (80^{\circ })=\sin (90^{\circ }-10^{\circ })=\cos (10^{\circ })$, we get:

$$\cos (10^{\circ })\cos (20^{\circ })\cos (40^{\circ })=\frac{\cos (10^{\circ })}{8\sin (10^{\circ })}=\frac{1}{8}\cot (10^{\circ })$$

Step 4: Combine everything to find $\alpha$

Substitute this back into the expression for $E$:

$$E=\frac{\sqrt{3}}{8}\left(\frac{\sin (80^{\circ })}{8\sin (10^{\circ })}\right)$$

$$E=\frac{\sqrt{3}}{64}\frac{\cos (10^{\circ })}{\sin (10^{\circ })}$$

We want to compare this with the right-hand side structure given in the question: $\alpha +\frac{\sqrt{3}}{16}\cos (10^{\circ })$. Let's manipulate $E$ to force this term out:

$$E=\frac{\sqrt{3}}{64}\cot (10^{\circ })$$

Alternatively, let's look at the original expression directly with a different grouping to isolate $\alpha$:

We know from Step 2 that:

$$E=\frac{\sqrt{3}}{8}\cos (10^{\circ })\cos (20^{\circ })\cos (40^{\circ })$$

Using the identity $\cos (A)\cos (B)=\frac{1}{2}[\cos (A+B)+\cos (A-B)]$ for $\cos (20^{\circ })\cos (40^{\circ })$:

$$\cos (20^{\circ })\cos (40^{\circ })=\frac{1}{2}[\cos (60^{\circ })+\cos (20^{\circ })]=\frac{1}{2}\left[\frac{1}{2}+\cos (20^{\circ })\right]=\frac{1}{4}+\frac{1}{2}\cos (20^{\circ })$$

Substitute this back into $E$:

$$E=\frac{\sqrt{3}}{8}\cos (10^{\circ })\left(\frac{1}{4}+\frac{1}{2}\cos (20^{\circ })\right)$$

$$E=\frac{\sqrt{3}}{32}\cos (10^{\circ })+\frac{\sqrt{3}}{16}\cos (10^{\circ })\cos (20^{\circ })$$

Now, apply the product-to-sum formula to $\cos (10^{\circ })\cos (20^{\circ })$:

$$\cos (20^{\circ })\cos (10^{\circ })=\frac{1}{2}[\cos (30^{\circ })+\cos (10^{\circ })]=\frac{1}{2}\left[\frac{\sqrt{3}}{2}+\cos (10^{\circ })\right]=\frac{\sqrt{3}}{4}+\frac{1}{2}\cos (10^{\circ })$$

Substitute this value into the expression:

$$E=\frac{\sqrt{3}}{32}\cos (10^{\circ })+\frac{\sqrt{3}}{16}\left(\frac{\sqrt{3}}{4}+\frac{1}{2}\cos (10^{\circ })\right)$$

$$E=\frac{\sqrt{3}}{32}\cos (10^{\circ })+\frac{3}{64}+\frac{\sqrt{3}}{32}\cos (10^{\circ })$$

$$E=\frac{3}{64}+\left(\frac{\sqrt{3}}{32}+\frac{\sqrt{3}}{32}\right)\cos (10^{\circ })$$

$$E=\frac{3}{64}+\frac{\sqrt{3}}{16}\cos (10^{\circ })$$

Step 5: Compare with the given equation

The given equation is:

$$E=\alpha +\frac{\sqrt{3}}{16}\cos (10^{\circ })$$

By comparing both sides, we get:

$$\alpha =\frac{3}{64}$$

Step 6: Calculate $3{\alpha }^{-1}$

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

Now, multiply by $3$:

$$3{\alpha }^{-1}=3\times \frac{64}{3}=64$$