A) \[{{A}_{n+1}}={{A}_{n}},{{B}_{n+1}}-{{B}_{n}}={{A}_{n+1}}\]
B) \[{{B}_{n+1}}={{B}_{n}}\]
C) \[{{A}_{n+1}}-{{A}_{n}}={{B}_{n+1}}\]
D) None of these
Correct Answer: A
Solution :
[a] We have \[{{A}_{n+1}}-{{A}_{n}}=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin (2n+1)x-\sin (2n-1)x}{\sin x}dx}\] \[=\int\limits_{0}^{\frac{\pi }{2}}{\frac{2\cos 2nx\,\,\sin \,\,x}{\sin x}dx=2\int\limits_{0}^{\frac{\pi }{2}}{\cos 2\,\,n\pi \,\,dx=0}}\] Again \[{{B}_{n+1}}-{{B}_{n}}=\int\limits_{0}^{\frac{\pi }{2}}{\frac{{{\sin }^{2}}(n+1)x-{{\sin }^{2}}nx}{{{\sin }^{2}}x}dx}\] \[=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin (2n+1)x\,\,\sin \,\,x}{{{\sin }^{2}}x}dx={{A}_{n+1}}}\]
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