यदि θ न्यूनकोण है और cos² θ - sin² θ = -1/2 है, तो [cos² 2θ + cot² (θ/2)] का मान क्या है?

Steps to Solve

1. Simplify the given equation using trigonometric identity

We are given that $\cos^2 \theta - \sin^2 \theta = -\frac{1}{2}$. Using the trigonometric identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$, we can rewrite the given equation as $$ \cos 2\theta = -\frac{1}{2} $$

2. Solve for $\theta$

Since $\theta$ is an acute angle (न्यूनकोण), $0 < \theta < \frac{\pi}{2}$, which implies $0 < 2\theta < \pi$. We have $\cos 2\theta = -\frac{1}{2}$. The angle $2\theta$ in the interval $(0, \pi)$ that satisfies this condition is $2\theta = \frac{2\pi}{3}$. Therefore, $\theta = \frac{\pi}{3}$.

3. Calculate $\cos^2 2\theta$ $2\theta = \frac{2\pi}{3}$, so $4\theta = \frac{4\pi}{3}$. $$ \cos^2 2\theta = \cos^2 \left(\frac{2\pi}{3}\right) = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} $$

4. Calculate $\cot^2 (\theta/2)$

Since $\theta = \frac{\pi}{3}$, we have $\frac{\theta}{2} = \frac{\pi}{6}$. Therefore, $$ \cot^2 \left(\frac{\theta}{2}\right) = \cot^2 \left(\frac{\pi}{6}\right) = (\sqrt{3})^2 = 3 $$

5. Evaluate the expression $\cos^2 2\theta + \cot^2 (\theta/2)$

$$ \cos^2 2\theta + \cot^2 \left(\frac{\theta}{2}\right) = \frac{1}{4} + 3 = \frac{1}{4} + \frac{12}{4} = \frac{13}{4} $$