Pinnacle Q.52. If \( \frac{1}{\operatorname{cosec} \theta-1}+\frac{1}{\operatorname{cosec} \theta+1}=2 \) \( \sec \theta, 0^{\circ}<\theta<90^{\circ} \), then the value of \( (\cot \theta+\cos \theta) \) is: \( \mathrm{SSC} \operatorname{CGL} 07 / 06 / 2019 \) (Evening)

To solve the equation \[ \frac{1}{\operatorname{cosec} \theta - 1} + \frac{1}{\operatorname{cosec} \theta + 1} = 2, \] we start by letting \( x = \operatorname{cosec} \theta \). Thus, the equation becomes: \[ \frac{1}{x - 1} + \frac{1}{x + 1} = 2. \] Next, we find a common denominator for the left-hand side: \[ \frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)} = 2. \] This simplifies to: \[ \frac{2x}{x^2 - 1} = 2. \] Now, we can cross-multiply: \[ 2x = 2(x^2 - 1). \] Expanding the right-hand side gives: \[ 2x = 2x^2 - 2. \] Rearranging this equation leads to: \[ 2x^2 - 2x - 2 = 0. \] Dividing the entire equation by 2 simplifies it to: \[ x^2 - x - 1 = 0. \] Now, we can use the quadratic formula to solve for \( x \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2}. \] Since \( x = \operatorname{cosec} \theta \) must be positive, we take the positive root: \[ \operatorname{cosec} \theta = \frac{1 + \sqrt{5}}{2}. \] Now, we can find \( \sin \theta \): \[ \sin \theta = \frac{1}{\operatorname{cosec} \theta} = \frac{2}{1 + \sqrt{5}}. \] Next, we rationalize the denominator: \[ \sin \theta = \frac{2(1 - \sqrt{5})}{(1 + \sqrt{5})(1 - \sqrt{5})} = \frac{2(1 - \sqrt{5})}{1 - 5} = \frac{2(1 - \sqrt{5})}{-4} = -\frac{1 - \sqrt{5}}{2}. \] Now, we can find \( \cos \theta \) using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(-\frac{1 - \sqrt{5}}{2}\right)^2. \] Calculating \( \sin^2 \theta \): \[ \sin^2 \theta = \frac{(1 - \sqrt{5})^2}{4} = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}. \] Thus, \[ \cos^2 \theta = 1 - \frac{3 - \sqrt{5}}{2} = \frac{2 - (3 - \sqrt{5})}{2} = \frac{-1 + \sqrt{5}}{2}. \] Now, we can find \( \cot \theta \): \[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{\frac{-1 + \sqrt{5}}{2}}}{-\frac{1 - \sqrt{5}}{2}} = -\frac{\sqrt{-1 + \sqrt{5}}}{1 - \sqrt{5}}. \] Finally, we need to find \( \cot \theta + \cos \theta \). However, since this is getting complex, let's calculate \( \cot \theta + \cos \theta \) directly using the values we have: 1. \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) 2. \( \cos \theta = \sqrt{\frac{-1 + \sqrt{5}}{2}} \) Now, we can find \( \cot \theta + \cos \theta \): \[ \cot \theta + \cos \theta = -\frac{\sqrt{-1 + \sqrt{5}}}{1 - \sqrt{5}} + \sqrt{\frac{-1 + \sqrt{5}}{2}}. \] This expression can be simplified further, but let's calculate the final value directly. Now, let's calculate \( \cot \theta + \cos \theta \) using the known values. We can use the values of \( \sin \theta \) and \( \cos \theta \) to find \( \cot \theta + \cos \theta \): \[ \cot \theta + \cos \theta = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta} = \frac{1}{\sin \theta}. \] Thus, we can find the final value of \( \cot \theta + \cos \theta \): \[ \cot \theta + \cos \theta = \frac{1 + \sqrt{5}}{2}. \] Now, let's calculate the final value: \[ \cot \theta + \cos \theta = \frac{1 + \sqrt{5}}{2}. \] Thus, the final answer is: \[ \boxed{1}. \]