Question:

Step 1: Analyze Condition (II) If $\cos\theta$, $\sec\theta$, $\cot\theta$ are three positive numbers in H.P., then their reciprocals, $\sec\theta$, $\cos\theta$, $\tan\theta$, are in A.P. This implies the relation: $$2\cos\theta = \sec\theta + \tan\theta$$ Step 2: Formulate the trigonometric equation Substitute $\sec\theta = \frac{1}{\cos\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$ into the equation: $$2\cos\theta = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}$$ Multiply both sides by $\cos\theta$ (note that $\cos\theta \neq 0$ since $\sec\theta$ is defined and positive): $$2\cos^2\theta = 1 + \sin\theta$$ Using the identity $\cos^2\theta = 1 - \sin^2\theta$: $$2(1 - \sin^2\theta) = 1 + \sin\theta$$ $$2 - 2\sin^2\theta = 1 + \sin\theta$$ Rearrange the terms to form a quadratic equation in terms of $\sin\theta$: $$2\sin^2\theta + \sin\theta - 1 = 0$$ Step 3: Solve for $\sin\theta$ Factor the quadratic equation: $$(2\sin\theta - 1)(\sin\theta + 1) = 0$$ This yields two possible values for $\sin\theta$: $$\sin\theta = \frac{1}{2} \quad \text{or} \quad \sin\theta = -1$$ Step 4: Apply positivity constraints The problem states that $\cos\theta$, $\sec\theta$, and $\cot\theta$ are all positive numbers. For $\cos\theta > 0$ and $\cot\theta = \frac{\cos\theta}{\sin\theta} > 0$, it must be that $\sin\theta > 0$. Therefore, we must choose $\sin\theta = \frac{1}{2}$. The value $\sin\theta = -1$ would imply $\theta = \frac{3\pi}{2} + 2n\pi$, for which $\cos\theta = 0$. This would make $\sec\theta$ undefined and $\cot\theta = 0$, violating the condition that these terms are positive. Step 5: Determine $\cos\theta$ Since $\sin\theta = \frac{1}{2}$ and $\theta$ must be in the first quadrant (as $\sin\theta > 0$ and $\cos\theta > 0$), we find $\cos\theta$: $$\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$ Step 6: Calculate $\sin\theta + \cos\theta$ Using the determined values of $\sin\theta$ and $\cos\theta$: $$\sin\theta + \cos\theta = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}$$ This value corresponds to Column-2 (i). Step 7: Determine the general value of $\theta$ From $\sin\theta = \frac{1}{2}$ and $\cos\theta = \frac{\sqrt{3}}{2}$, the principal value of $\theta$ is $\frac{\pi}{6}$. The general solution for $\theta$ is: $$\theta = 2n\pi + \frac{\pi}{6}, \quad \text{where } n \in \mathbb{Z}$$ This general value corresponds to Column-3 (Q).