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Trigonometric Equations

If x+12/x3 + ...

Question

If $\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}$ , then $c o s e c^{- 1} (\frac{1}{A}) + c o t^{- 1} (\frac{1}{B}) + s e c^{- 1} C =$

A

$\frac{5 π}{6}$

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B

$0$

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C

$\frac{π}{6}$

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D

$\frac{π}{2}$

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Solution

The correct option is A
$\frac{5 π}{6}$

A = 1, B = 0, C = 2.

So $c o s e c^{- 1} 1 + 0 + s e c^{- 1} 2 = \frac{π}{2} + 0 + \frac{π}{3} = 5 \frac{π}{6}$

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Similar questions

\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1},

c o s e c^{- 1} (\frac{1}{A}) + c o t^{- 1} (\frac{1}{B}) + s e c^{- 1} C =

\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}

{csc}^{- 1} (\frac{1}{A}) + {cot}^{- 1} (\frac{1}{B}) + {sec}^{- 1} C =

\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}

{sin}^{- 1} A + {tan}^{- 1} B + {sec}^{- 1} C =

\frac{(x - 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}

A = . . ., B = . . ., C = . . .

\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}

{sin}^{- 1} (\frac{A}{C}) =

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