A+B = π/2 হলে cos²A - cos²B এর মান-

Explanation:

Another Explanation (5): দেওয়া আছে, \(A + B = \frac{\pi}{2}\) তাহলে, \(A = \frac{\pi}{2} - B\) এখন, \(\cos^2 A - \cos^2 B = \cos^2 (\frac{\pi}{2} - B) - \cos^2 B\) আমরা জানি, \(\cos (\frac{\pi}{2} - \theta) = \sin \theta\) সুতরাং, \(\cos^2 (\frac{\pi}{2} - B) = \sin^2 B\) তাহলে, \(\cos^2 A - \cos^2 B = \sin^2 B - \cos^2 B\) আমরা জানি, \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\) সুতরাং, \(-\cos 2B = \sin^2 B - \cos^2 B\) অতএব, \(\cos^2 A - \cos^2 B = -\cos 2B\) আবার, \(A + B = \frac{\pi}{2}\) সুতরাং, \(2A + 2B = \pi\) বা, \(2A = \pi - 2B\) বা, \(2B = \pi - 2A\) এখন, \(-\cos 2B = -\cos (\pi - 2A)\) আমরা জানি, \(\cos (\pi - \theta) = -\cos \theta\) সুতরাং, \(-\cos (\pi - 2A) = -(-\cos 2A) = \cos 2A\) তাহলে, \(\cos^2 A - \cos^2 B = \cos 2A\) আমরা জানি, \(\cos 2A = \cos (A + A) = \cos ((\frac{\pi}{2} - B) + A) = \cos (\frac{\pi}{2} + (A - B))\) আবার, \(\cos (\frac{\pi}{2} + \theta) = -\sin \theta\) সুতরাং, \(\cos (\frac{\pi}{2} + (A - B)) = -\sin (A - B) = \sin (B - A)\) অতএব, \(\cos^2 A - \cos^2 B = \sin (B - A)\) 🥳🎉