(1+sinx)/(1-sinx)=? 

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

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

ব্যাখ্যা:

আমরা জানি, \(\sin x = \cos\left(\frac{\pi}{2} - x\right)\)। সুতরাং,

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

আমরা আরও জানি, \(1+\cos 2\theta = 2\cos^2\theta\) এবং \(1-\cos 2\theta = 2\sin^2\theta\)। তাই, \(2\theta = \frac{\pi}{2}-x\) ধরলে, \(\theta = \frac{\pi}{4}-\frac{x}{2}\) হয়।

সুতরাং, \(1+\cos\left(\frac{\pi}{2}-x\right) = 2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\) এবং \(1-\cos\left(\frac{\pi}{2}-x\right) = 2\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\)

তাহলে, \(\frac{1+\cos\left(\frac{\pi}{2}-x\right)}{1-\cos\left(\frac{\pi}{2}-x\right)} = \frac{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} = \cot^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\)

এখন, \(\cot\left(\frac{\pi}{4}-\frac{x}{2}\right) = \frac{1}{\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}\)

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

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

তাহলে, \(\cot\left(\frac{\pi}{4}-\frac{x}{2}\right) = \frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\)

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

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

অতএব, \(\frac{1+\sin x}{1-\sin x} = \cot^2\left(\frac{\pi}{4}-\frac{x}{2}\right) = \tan^2\left(\frac{\pi}{4}+\frac{x}{2}\right)\)

✅ প্রমাণিত।