NDA 2026 — Mathematics PYQ

To solve this, we will use the Product-to-Sum trigonometric formula:

$$2\sin A\sin B=\cos (A-B)-\cos (A+B)$$

Let $A=\frac{\theta }{2}$ and $B=\frac{3\theta }{2}$. The expression becomes:

$$\sin \left(\frac{\theta }{2}\right)\sin \left(\frac{3\theta }{2}\right)=\frac{1}{2}\left[\cos \left(\frac{\theta }{2}-\frac{3\theta }{2}\right)-\cos \left(\frac{\theta }{2}+\frac{3\theta }{2}\right)\right]$$

Step 1: Simplify the arguments inside the cosine function

• $A-B=\frac{\theta -3\theta }{2}=-\frac{2\theta }{2}=-\theta$

• $A+B=\frac{\theta +3\theta }{2}=\frac{4\theta }{2}=2\theta$

Substituting these back:

$$\text{Expression}=\frac{1}{2}[\cos (-\theta )-\cos (2\theta )]$$

Since $\cos (-\theta )=\cos \theta$:

$$\text{Expression}=\frac{1}{2}[\cos \theta -\cos (2\theta )]$$

Step 2: Use the double-angle formula for cosine

We know that $\cos (2\theta )=2{\cos }^2\theta -1$. Substitute this into the equation:

$$\text{Expression}=\frac{1}{2}[\cos \theta -(2{\cos }^2\theta -1)]$$

$$\text{Expression}=\frac{1}{2}[\cos \theta -2{\cos }^2\theta +1]$$

Step 3: Substitute the given value $\cos \theta =\frac{1}{3}$

$$\text{Expression}=\frac{1}{2}\left[\frac{1}{3}-2{\left(\frac{1}{3}\right)}^2+1\right]$$

$$\text{Expression}=\frac{1}{2}\left[\frac{1}{3}-2\left(\frac{1}{9}\right)+1\right]$$

$$\text{Expression}=\frac{1}{2}\left[\frac{1}{3}-\frac{2}{9}+1\right]$$

Find a common denominator (which is 9):

$$\text{Expression}=\frac{1}{2}\left[\frac{3}{9}-\frac{2}{9}+\frac{9}{9}\right]$$

$$\text{Expression}=\frac{1}{2}\left[\frac{10}{9}\right]=\frac{5}{9}$$

Conclusion: The correct option is (a) $\frac{5}{9}$.