(1+sinx)/(1- sinx) =? 

প্রশ্ন: \(\frac{1+\sin x}{1-\sin x} = ?\)

উত্তর: \(\tan^2(\frac{\pi}{4}+\frac{x}{2})\)

ব্যাখ্যা:

আমরা জানি, \(\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}\) এবং \(1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}\)

সুতরাং, \(1 + \sin x = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2\)

এবং \(1 - \sin x = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} - \sin \frac{x}{2})^2\)

অতএব, \(\frac{1+\sin x}{1-\sin x} = \frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})^2} = \left( \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \right)^2\)

এখন, লব ও হরকে \(\cos \frac{x}{2}\) দিয়ে ভাগ করে পাই,

\(\left( \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} \right)^2\)

আমরা জানি, \(\tan \frac{\pi}{4} = 1\). সুতরাং,

\(\left( \frac{\tan \frac{\pi}{4} + \tan \frac{x}{2}}{1 - \tan \frac{\pi}{4} \tan \frac{x}{2}} \right)^2\)

আমরা জানি, \(\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)

সুতরাং, \(\left( \tan (\frac{\pi}{4} + \frac{x}{2}) \right)^2 = \tan^2 (\frac{\pi}{4} + \frac{x}{2})\) 🎉