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Let S = _ r=1 ³ cosec ( 8 + (r-1) 4 ) cosec ( 8 + r 4 ) . If S = a + b 2 where a and b are rational numbers, then the quadratic equation whose roots are (a+b) and (a-b) is :

Options

  1. Ax^2 - 8x + 12 = 0
  2. Bx^2 - 4x = 0
  3. Cx^2 + 8x + 12 = 0
  4. Dx^2 - 4x - 12 = 0

Correct answer

A. x^2 - 8x + 12 = 0

Step-by-step solution

The given sum is S = _ r=1 ³ 1 ( 8 + (r-1) 4 ) ( 8 + r 4 ) . The difference between the angles in the denominator is 4 . Multiply and divide the entire sum by ( 4 ) : S = 1 ( 4 ) _ r=1 ³ ( ( 8 + r 4 ) - ( 8 + (r-1) 4 ) ) ( 8 + (r-1) 4 ) ( 8 + r 4 ) Using (A - B) = A B - A B , the sum becomes a difference of cotangents: S = 2 _ r=1 ³ [ ( 8 + (r-1) 4 ) - ( 8 + r 4 ) ] This is a telescoping series. Expanding it yields: S = 2 [ ( 8 - 3 8 ) + ( 3 8 - 5 8 ) + ( 5 8 - 7 8 ) ] S = 2 [ ( 8 ) - ( 7 8 ) ] Since ( 7 8 ) = ( -

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