If x-12-x3-x-A-x-B x-C-x2-1- then cosec-11-A- -11-B- -1 C-

(x+1)^2x^3+x=Ax+ Bx+Cx^2+1 nbsp; nbsp; ⇒(x+1)^2x^3+x=A(x^2+1)+x(Bx+C)x(x^2+1) Comparing the numerators, we get nbsp; A=1, B=0, C=2 nbsp; nbsp;∴ cosec ...

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Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

If x+12/x3+x=...

Question

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

\frac{5 π}{6}

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\frac{π}{6}

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\frac{π}{2}

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Solution

The correct option is A $\frac{5 π}{6}$
$\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}$
$\Rightarrow \frac{(x + 1)^{2}}{x^{3} + x} = \frac{A (x^{2} + 1) + x (B x + C)}{x (x^{2} + 1)}$
Comparing the numerators, we get
$A = 1, B = 0, C = 2$
$∴ c o s e c^{- 1} (\frac{1}{A}) + c o t^{- 1} (\frac{1}{B}) + s e c^{- 1} C$
$= \frac{π}{2} + 0 + \frac{π}{3} = \frac{5 π}{6}$

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