যদি sinA+cosA=sinB+cosB তবে, A+B=?

Explanation:

Another Explanation (5): দেওয়া আছে, sinA + cosA = sinB + cosB উভয়পক্ষকে \(\sqrt{2}\) দিয়ে ভাগ করে পাই, \(\frac{1}{\sqrt{2}}\)sinA + \(\frac{1}{\sqrt{2}}\)cosA = \(\frac{1}{\sqrt{2}}\)sinB + \(\frac{1}{\sqrt{2}}\)cosB cos(\(\frac{\pi}{4}\))sinA + sin(\(\frac{\pi}{4}\))cosA = cos(\(\frac{\pi}{4}\))sinB + sin(\(\frac{\pi}{4}\))cosB sin(A+\(\frac{\pi}{4}\)) = sin(B+\(\frac{\pi}{4}\)) সুতরাং, A+\(\frac{\pi}{4}\) = B+\(\frac{\pi}{4}\) অথবা, A+\(\frac{\pi}{4}\) = \(\pi\) - (B+\(\frac{\pi}{4}\)) ১ম ক্ষেত্র: A+\(\frac{\pi}{4}\) = B+\(\frac{\pi}{4}\) => A = B (যা সম্ভব নয়, কারণ A ≠ B) 😒 ২য় ক্ষেত্র: A+\(\frac{\pi}{4}\) = \(\pi\) - (B+\(\frac{\pi}{4}\)) => A+\(\frac{\pi}{4}\) = \(\pi\) - B - \(\frac{\pi}{4}\) => A + B = \(\pi\) - \(\frac{\pi}{4}\) - \(\frac{\pi}{4}\) => A + B = \(\pi\) - \(\frac{\pi}{2}\) => A + B = \(\frac{\pi}{2}\) 🎉 অতএব, A + B = \(\frac{\pi}{2}\) 🙏