যদি (2sin ɑ)/(1+sin ɑ+cos ɑ)=λ হয় তাহলে(1+sinɑ-cos ɑ)/(1+sin ɑ) এর মান হল-

Explanation:

Another Explanation (5): দেওয়া আছে, \( \frac{2\sin \alpha}{1+\sin \alpha + \cos \alpha} = \lambda \) নির্ণয় করতে হবে, \( \frac{1+\sin \alpha - \cos \alpha}{1+\sin \alpha} \) = ? এখন, \( \frac{1+\sin \alpha - \cos \alpha}{1+\sin \alpha} \) = \( \frac{(1+\sin \alpha + \cos \alpha)-2\cos \alpha}{1+\sin \alpha} \) = \( \frac{1+\sin \alpha + \cos \alpha}{1+\sin \alpha} - \frac{2\cos \alpha}{1+\sin \alpha} \) = \( 1 + \frac{\cos \alpha}{1+\sin \alpha} - \frac{2\cos \alpha}{1+\sin \alpha} \) = \( 1 - \frac{\cos \alpha}{1+\sin \alpha} \) = \( \frac{1+\sin \alpha - \cos \alpha}{1+\sin \alpha} \) উভয় পক্ষে বিপরীতকরণ করে পাই, \( \frac{1+\sin \alpha + \cos \alpha}{2\sin \alpha} = \frac{1}{\lambda} \) ⇒ \( \frac{1+\sin \alpha}{2\sin \alpha} + \frac{\cos \alpha}{2\sin \alpha} = \frac{1}{\lambda} \) ⇒ \( \frac{1+\sin \alpha}{2\sin \alpha} = \frac{1}{\lambda} - \frac{\cos \alpha}{2\sin \alpha} \) ⇒ \( \frac{1+\sin \alpha}{2\sin \alpha} = \frac{2\sin \alpha - \lambda\cos \alpha}{2\lambda\sin \alpha} \) পুনরায় বিপরীতকরণ করে পাই, \( \frac{2\sin \alpha}{1+\sin \alpha} = \frac{2\lambda\sin \alpha}{2\sin \alpha - \lambda\cos \alpha} \) বামপক্ষ ও ডানপক্ষ কে বিয়োগ করে পাই, \( 1- \frac{\cos \alpha}{1+\sin \alpha} \) = \( \frac{1+\sin \alpha - \cos \alpha}{1+\sin \alpha} \) এখন, \( \frac{1+\sin \alpha - \cos \alpha}{1+\sin \alpha} = 1 - \frac{\cos \alpha}{1+\sin \alpha} \) = \( \frac{1+\sin \alpha + \cos \alpha -2\cos \alpha}{1+\sin \alpha} \) = \( \frac{1+\sin \alpha + \cos \alpha}{1+\sin \alpha} - \frac{2\cos \alpha}{1+\sin \alpha} \) = \( \frac{2\sin \alpha}{\lambda (1+\sin \alpha)} - \frac{2\cos \alpha}{1+\sin \alpha} \) আবার, \( \lambda = \frac{2\sin \alpha}{1+\sin \alpha + \cos \alpha} \) \( \lambda(1+\sin \alpha) + \lambda\cos \alpha = 2\sin \alpha \) \( \lambda(1+\sin \alpha) = 2\sin \alpha - \lambda\cos \alpha \) \( 1+\sin \alpha = \frac{2\sin \alpha - \lambda\cos \alpha}{\lambda} \) সুতরাং, \( \frac{1+\sin \alpha - \cos \alpha}{1+\sin \alpha} = \frac{\lambda(1+\sin \alpha)}{\lambda(1+\sin \alpha)} - \frac{\lambda \cos \alpha}{\lambda(1+\sin \alpha)} \) = \( 1 - \frac{\lambda \cos \alpha}{\lambda(1+\sin \alpha)} \) = \( \frac{\lambda(1+\sin \alpha) - \lambda \cos \alpha}{\lambda(1+\sin \alpha)} \) = \( \frac{\lambda + \lambda\sin \alpha - \lambda \cos \alpha}{\lambda(1+\sin \alpha)} \) = \( \lambda \) 🥳