If x - 12x3 - x - Ax - Bx - Cx2 - 1- then -1 1A - -1 1B - -1C -

The correct option is A 5π6 We have, #160; (x + 1)^2x^3 + x = Ax + Bx + Cx^2 + 1=A(x^2+1)+x(Bx+C)x(x^2+1) , #160; ⇒ (x+1)^2=A(x^2+1)+x(Bx+C) ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If x + 12x3...

Question

If $\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}$ , then ${csc}^{- 1} (\frac{1}{A}) + {cot}^{- 1} (\frac{1}{B}) + {sec}^{- 1} C =$ ____

\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 $\frac{5 π}{6}$
We have, $\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1} = \frac{A (x^{2} + 1) + x (B x + C)}{x (x^{2} + 1)}$ ,
$\Rightarrow (x + 1)^{2} = A (x^{2} + 1) + x (B x + C)$
$\Rightarrow x^{2} + 2 x + 1 = (A + B) x^{2} + C x + A$
Equate the corresponding coefficients
$\Rightarrow A = 1, C = 2, A + B = 1 \Rightarrow B = 0$
Therefore,
${csc}^{- 1} (\frac{1}{A}) + {cot}^{- 1} (\frac{1}{B}) + {sec}^{- 1} C = {csc}^{- 1} 1 + {cot}^{- 1} (\infty) + {sec}^{- 1} 2$
$= \frac{π}{2} + 0 + \frac{π}{3} = \frac{5 π}{6}$

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If (x+1)2x3+x=Ax+Bx+Cx2+1, then csc−1(1A)+cot−1(1B)+sec−1C= ____

The correct option is A 5π6We have, (x+1)2x3+x=Ax+Bx+Cx2+1=A(x2+1)+x(Bx+C)x(x2+1), ⇒(x+1)2=A(x2+1)+x(Bx+C)⇒x2+2x+1=(A+B)x2+Cx+AEquate the corresponding coefficients ⇒A=1,C=2,A+B=1⇒B=0Therefore,csc−1(1A)+cot−1(1B)+sec−1C=csc−11+cot−1(∞)+sec−12=π2+0+π3=5π6

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