if g(theta)=(1-tantheta)/(1+tantheta), g(π/4-theta)=?  

Explanation:

Another Explanation (5): দেওয়া আছে, \( g(\theta) = \frac{1 - \tan\theta}{1 + \tan\theta} \) আমাদের \( g\left(\frac{\pi}{4} - \theta\right) \) এর মান নির্ণয় করতে হবে। \( g\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan\left(\frac{\pi}{4} - \theta\right)}{1 + \tan\left(\frac{\pi}{4} - \theta\right)} \) আমরা জানি, \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\) সুতরাং, \(\tan\left(\frac{\pi}{4} - \theta\right) = \frac{\tan\frac{\pi}{4} - \tan\theta}{1 + \tan\frac{\pi}{4} \tan\theta}\) যেহেতু \(\tan\frac{\pi}{4} = 1\), তাই \(\tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan\theta}{1 + \tan\theta}\) এখন, \( g\left(\frac{\pi}{4} - \theta\right) \) এর মান বসাই: \( g\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \frac{1 - \tan\theta}{1 + \tan\theta}}{1 + \frac{1 - \tan\theta}{1 + \tan\theta}} \) \( = \frac{\frac{(1 + \tan\theta) - (1 - \tan\theta)}{1 + \tan\theta}}{\frac{(1 + \tan\theta) + (1 - \tan\theta)}{1 + \tan\theta}} \) \( = \frac{1 + \tan\theta - 1 + \tan\theta}{1 + \tan\theta + 1 - \tan\theta} \) \( = \frac{2\tan\theta}{2} \) \( = \tan\theta \) অতএব, \( g\left(\frac{\pi}{4} - \theta\right) = \tan\theta \) 🎉🎉