A+B+C = pi হয় তবে cos²A +cos²B +cos²C + 2cosAcosBcosC =?

Explanation:

Another Explanation (5): bài giải: যেহেতু \(A+B+C = \pi\), প্রমাণ করতে হবে: \(\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B \cos C = 1\). বামপক্ষ=\(\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B \cos C\) =\(\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B \cos (\pi - (A+B))\) =\(\cos^2 A + \cos^2 B + \cos^2 C - 2\cos A \cos B \cos (A+B)\) =\(\cos^2 A + \cos^2 B + \cos^2 C - 2\cos A \cos B (\cos A \cos B - \sin A \sin B)\) =\(\cos^2 A + \cos^2 B + \cos^2 C - 2\cos^2 A \cos^2 B + 2\cos A \cos B \sin A \sin B\) =\(\cos^2 A + \cos^2 B + \cos^2 C - 2\cos^2 A \cos^2 B + 2\cos A \sin A \cos B \sin B\) =\(\cos^2 A + \cos^2 B + \cos^2 C - 2\cos^2 A \cos^2 B + \frac{1}{2}(2\sin A \cos A)(2\sin B \cos B)\) =\(\cos^2 A + \cos^2 B + \cos^2 C - 2\cos^2 A \cos^2 B + \frac{1}{2}\sin 2A \sin 2B\) এখন, \(C = \pi - (A+B)\) হওয়ায়, \(\cos^2 C = \cos^2 (\pi - (A+B)) = \cos^2 (A+B)\). সুতরাং, \(\cos^2 A + \cos^2 B + \cos^2 (A+B) - 2\cos^2 A \cos^2 B + 2\cos A \cos B \sin A \sin B\) \(\cos^2 A + \cos^2 B + (\cos A \cos B - \sin A \sin B)^2 - 2\cos^2 A \cos^2 B + 2\cos A \cos B \sin A \sin B\) \(\cos^2 A + \cos^2 B + \cos^2 A \cos^2 B - 2\cos A \cos B \sin A \sin B + \sin^2 A \sin^2 B - 2\cos^2 A \cos^2 B + 2\cos A \cos B \sin A \sin B\) \(\cos^2 A + \cos^2 B - \cos^2 A \cos^2 B + \sin^2 A \sin^2 B\) \(\cos^2 A + \cos^2 B - \cos^2 A \cos^2 B + (1-\cos^2 A)(1-\cos^2 B)\) \(\cos^2 A + \cos^2 B - \cos^2 A \cos^2 B + 1 - \cos^2 A - \cos^2 B + \cos^2 A \cos^2 B\) = \(1\) = ডানপক্ষ 🥳 সুতরাং, \(\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B \cos C = 1\). প্রমাণিত।✅