`IfA_n=int_0^(pi/2)(sin(2n-1)x)/(sinx)dx ,b_n=int_0^(pi/2)((sinn x)/(sinx))^2dxforn in N ,` Then `A_(n+1)=A_n` (b) `B_(n+1)=B_n` `A_(n+1)-A_n=B_(n+1)`

Monjur Alahi

`IfA_n=int_0^(pi/2)(sin(2n-1)x)/(sinx)dx ,b_n=int_0^(pi/2)((sinn x)/(sinx))^2dxforn in N ,` Then `A_(n+1)=A_n` (b) `B_(n+1)=B_n` `A_(n+1)-A_n=B_(n+1)` (d) `B_(n+1)-B_n=A_(n+1)`
A. `A_(n+1)=A_(n)`
B. `B_(n+1)=B_(n)`
C. `A_(n+1)-A_(n)=B_(n+1)`
D. `B_(n+1)-B_(n)=A_(n+1)`


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Salim Ahmed Kawsar

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Correct Answer - A::D
`A_(n+1)-A_(n)=int_(0)^(pi//2)(sin(2n+1)xsin(2n-1)x)/(sinx) dx`
`=int_(0)^(pi//2)2 cos 2nx dx=0`
or `A_(n+1)=A_(n)`
`B_(n+1)-B_(n)=int_(0)^(pi//2)(sin^(2)(n+1)x-sin^(2)nx)/(sin^(2)x)dx`
`=int_(0)^(pi//2) (sin(2n+1)x)/(sinx)dx=A_(n+1)`

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