costheta=(acosφ-b)/(a-bcosφ) হলে (tan(theta/2))/(tan(φ/2)) এর মান কোনটি? 

প্রশ্নানুসারে, \( \cos\theta = \frac{a\cos\phi - b}{a - b\cos\phi} \) 🤔

আমরা জানি, \( \cos\theta = \frac{1 - \tan^2(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} \) এবং \( \cos\phi = \frac{1 - \tan^2(\frac{\phi}{2})}{1 + \tan^2(\frac{\phi}{2})} \) 🤗

সুতরাং, \( \frac{1 - \tan^2(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} = \frac{a\frac{1 - \tan^2(\frac{\phi}{2})}{1 + \tan^2(\frac{\phi}{2})} - b}{a - b\frac{1 - \tan^2(\frac{\phi}{2})}{1 + \tan^2(\frac{\phi}{2})}} \) 🤓

এখন, \( \frac{1 - \tan^2(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} = \frac{a(1 - \tan^2(\frac{\phi}{2})) - b(1 + \tan^2(\frac{\phi}{2}))}{a(1 + \tan^2(\frac{\phi}{2})) - b(1 - \tan^2(\frac{\phi}{2}))} \) 🤩

সরলীকরণ করে পাই, \( \frac{1 - \tan^2(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} = \frac{a - a\tan^2(\frac{\phi}{2}) - b - b\tan^2(\frac{\phi}{2})}{a + a\tan^2(\frac{\phi}{2}) - b + b\tan^2(\frac{\phi}{2})} \) 😎

আরও সরলীকরণ করে পাই, \( \frac{1 - \tan^2(\frac{\theta}{2})}{1 + \tan^2(\frac{\theta}{2})} = \frac{(a - b) - (a + b)\tan^2(\frac{\phi}{2})}{(a - b) + (a + b)\tan^2(\frac{\phi}{2})} \) 🧐

যোগ ভাগ প্রক্রিয়া করে পাই, \( \frac{1 - \tan^2(\frac{\theta}{2}) + 1 + \tan^2(\frac{\theta}{2})}{1 - \tan^2(\frac{\theta}{2}) - 1 - \tan^2(\frac{\theta}{2})} = \frac{(a - b) - (a + b)\tan^2(\frac{\phi}{2}) + (a - b) + (a + b)\tan^2(\frac{\phi}{2})}{(a - b) - (a + b)\tan^2(\frac{\phi}{2}) - (a - b) - (a + b)\tan^2(\frac{\phi}{2})} \) 🤯

আরও সরলীকরণ করে পাই, \( \frac{2}{-2\tan^2(\frac{\theta}{2})} = \frac{2(a - b)}{-2(a + b)\tan^2(\frac{\phi}{2})} \) 😮

অতএব, \( \frac{1}{\tan^2(\frac{\theta}{2})} = \frac{a - b}{(a + b)\tan^2(\frac{\phi}{2})} \) 😴

সুতরাং, \( \frac{\tan^2(\frac{\theta}{2})}{\tan^2(\frac{\phi}{2})} = \frac{a + b}{a - b} \) 😌

অতএব, \( \frac{\tan(\frac{\theta}{2})}{\tan(\frac{\phi}{2})} = \sqrt{\frac{a + b}{a - b}} \) 🎉