If x3+x(x+1)2=xA+x2+1Bx+C, then sin−1A+tan−1B+sec−1C=

We have: x3+x(x+1)2=xA+x2+1Bx+C⇒x3+x(x+1)2=x3+xA(x2+1)+x(Bx+C)⇒x2+2x+1=Ax2+A+Bx2+Cx⇒A+B=1,C=2,A=1∴B=0⇒sin−1A+tan−1B+sec−1C=sin−11+tan−10+sec−12=2π+0+3π=65π

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If {(x+1)^2}{x^3+x}={A}{x}+{Bx+C}{x^2+1}, then ^{-1} A+ ^{-1}B

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

\frac{π}{2}

\frac{π}{6}

\frac{5 π}{6}

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Levi Discussed

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

, then

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

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Text solutionVerified

We have:

\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 ^{3} + x}

\Rightarrow x^{2} + 2 x + 1 = A x^{2} + A + B x^{2} + C x

\Rightarrow A + B = 1, C = 2, A = 1

∴ B = 0

\Rightarrow sin^{- 1} A + tan^{- 1} B + sec^{- 1} C = sin^{- 1} 1 + tan^{- 1} 0 + sec^{- 1} 2

= \frac{π}{2} + 0 + \frac{π}{3}

= \frac{5 π}{6}

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Question Text	If $\frac{( x + 1 ) ^{2}}{x ^{3} + x} = \frac{A}{x} + \frac{B x + C}{x ^{2} + 1}$ , then $sin^{- 1} A + tan^{- 1} B + sec^{- 1} C =$
Answer Type	Text solution:1
Upvotes	151

If x3+x(x+1)2​=xA​+x2+1Bx+C​, then sin−1A+tan−1B+sec−1C=

Text solutionVerified

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