sintheta+costheta=sqrt2  হলে, 0lethetalepi/2 ব্যবধিতে θ এর মান কত?

Explanation:

Another Explanation (5): দেওয়া আছে, \( \sin\theta + \cos\theta = \sqrt{2} \) এবং \( 0 \le \theta \le \frac{\pi}{2} \) উভয়পক্ষকে \(\sqrt{2}\) দিয়ে ভাগ করে পাই, \[ \frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta = 1 \] আমরা জানি, \( \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \) এবং \( \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \) সুতরাং, \[ \sin\theta\cos\frac{\pi}{4} + \cos\theta\sin\frac{\pi}{4} = 1 \] \[ \sin(\theta + \frac{\pi}{4}) = 1 \] আমরা জানি, \( \sin\frac{\pi}{2} = 1 \) সুতরাং, \[ \sin(\theta + \frac{\pi}{4}) = \sin\frac{\pi}{2} \] অতএব, \[ \theta + \frac{\pi}{4} = \frac{\pi}{2} \] \[ \theta = \frac{\pi}{2} - \frac{\pi}{4} \] \[ \theta = \frac{2\pi - \pi}{4} \] \[ \theta = \frac{\pi}{4} \] যেহেতু \( 0 \le \frac{\pi}{4} \le \frac{\pi}{2} \), তাই \(\theta = \frac{\pi}{4}\) গ্রহণযোগ্য। সুতরাং, \(\theta = \frac{\pi}{4}\) 🥳🎉