JEE Mathematics 2026 Question with Solution

Given: $\frac{\pi}{2} < \theta < \pi$ and $\cot \theta = -\frac{1}{2\sqrt{2}}$.

Find: The value of the given trigonometric expression.

From the solution, let

$$A = \frac{15\theta}{2}, \qquad B = 8\theta$$

Then the expression is

$$\sin A \cos B + \sin A \sin B + \cos A \cos B - \cos A \sin B$$

Group the terms as

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

Using identities,

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

and

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

So the expression becomes

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

Now,

$$A-B = \frac{15\theta}{2} - \frac{16\theta}{2} = -\frac{\theta}{2}$$

Hence,

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

Given $\cot \theta = -\frac{1}{2\sqrt{2}}$, we get

$$\tan \theta = -2\sqrt{2}$$

Since $\theta$ lies in quadrant II, $\cos \theta < 0$ and $\sin \theta > 0$. From the working,

$$\cos \theta = -\frac{1}{3}$$

Since $\theta$ is in quadrant II, $\frac{\theta}{2}$ lies in quadrant I. Therefore the positive square roots are taken:

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

Therefore,

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

So the correct option is C.

Given: $\cot \theta = -\frac{1}{2\sqrt{2}}$ with $\frac{\pi}{2} < \theta < \pi$.

Find: Simplify the expression using angle identities.

The given combination can be rewritten by pairing terms that match standard identities:

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

Thus the full expression becomes

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

With $A = \frac{15\theta}{2}$ and $B = 8\theta$,

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

Therefore,

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

Now use the half-angle formulas after obtaining $\cos \theta = -\frac{1}{3}$ from the given cotangent value. Since $\frac{\theta}{2}$ is in quadrant I,

$$\sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\cos\theta}{2}} = \sqrt{\frac{2}{3}}, \qquad \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1+\cos\theta}{2}} = \sqrt{\frac{1}{3}}$$

Substitute these values:

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

Hence the correct option is C.