Q.If \[ \frac{\sec \theta + \tan \theta}{\sec \theta - \tan \theta} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}, \] then what is the value of \( \theta \)? (a) 60° (b) 30° (c) 45° (d) 90°
Answer: A
\[ \frac{\sec \theta + \tan \theta}{\sec \theta - \tan \theta} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \\ \Rightarrow \frac{\sec \theta + \tan \theta + \sec \theta - \tan \theta}{\sec \theta + \tan \theta - \sec \theta + \tan \theta} \] \[ = \frac{2 + \sqrt{3} + 2 - \sqrt{3}}{2 + \sqrt{3} - 2 + \sqrt{3}} \\ \Rightarrow \frac{2\sec \theta}{2\tan \theta} = \frac{4}{2\sqrt{3}} \\ \Rightarrow \frac{\sec \theta}{\tan \theta} = \frac{2}{\sqrt{3}} \\ \Rightarrow \frac{\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{2}{\sqrt{3}} \\ \Rightarrow \frac{1}{\cos \theta} \times \frac{\cos \theta}{\sin \theta} = \frac{2}{\sqrt{3}} \\ \Rightarrow \csc \theta = \frac{2}{\sqrt{3}} = \csc 60^\circ \\ \Rightarrow \theta = 60^\circ \]
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