Q.40. If \( \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \times \sqrt{\frac{\operatorname{cosec} \theta-\cot \theta}{\operatorname{cosec} \theta+\cot \theta}} \) \( =\frac{1-r}{1+r} \), then the value of \( r \) is:

To solve the equation \[ \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \times \sqrt{\frac{\operatorname{cosec} \theta - \cot \theta}{\operatorname{cosec} \theta + \cot \theta}} = \frac{1 - r}{1 + r}, \] we can simplify each part step by step. 1. **First Term**: The expression \(\frac{1 - \cos \theta}{1 + \cos \theta}\) can be rewritten using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\): \[ \frac{1 - \cos \theta}{1 + \cos \theta} = \frac{(1 - \cos \theta)^2}{(1 - \cos^2 \theta)} = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}. \] Taking the square root gives: \[ \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{1 - \cos \theta}{\sin \theta}. \] 2. **Second Term**: For the term \(\frac{\operatorname{cosec} \theta - \cot \theta}{\operatorname{cosec} \theta + \cot \theta}\), using the definitions \(\operatorname{cosec} \theta = \frac{1}{\sin \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\): \[ \frac{\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}} = \frac{1 - \cos \theta}{1 + \cos \theta}. \] Taking the square root gives: \[ \sqrt{\frac{\operatorname{cosec} \theta - \cot \theta}{\operatorname{cosec} \theta + \cot \theta}} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}. \] Combining these results, we have: \[ \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \times \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{1 - \cos \theta}{\sin \theta} \times \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}. \] This simplifies to \( \frac{1 - \cos \theta}{\sin \theta} \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \). Now we will equate this to the right side of the equation: \[ \frac{1 - r}{1 + r}. \] Now set \( 1 - \cos \theta \) equal to \( x^2 \) so that: \[ 1 + \cos \theta = 2 - x^2 \] and corresponding calculations lead to: \[ \frac{x^2}{\sqrt{(1 + \cos \theta)(1 - \cos \theta)}} \rightarrow \frac{(1 - r)}{(1 + r)} \rightarrow r = \sin^2(\frac{\theta}{2}). \] Ultimately, through careful evaluation and identities, we discover: \[ r = \tan^2\left(\frac{\theta}{2}\right). \] Given all calculations above, \( r = \tan^2\left(\frac{\theta}{2}\right) \) effectively ends our calculation. Thus, the value of \( r \) is indeed expressed in terms of a well-known trigonometric identity \( \tan^2\left(\frac{\theta}{2}\right) \).