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Mathematics Question on Inverse Trigonometric Functions

If (x+1)2x3+x=Ax+Bx+Cx2+1\frac {(x+1)^2}{x^3+x}=\frac {A}{x}+\frac {Bx+C}{x^2+1}, then sin⁡−1 A+tan⁡−1 B+sec−1 C\sin^{-1} \,A + \tan ^{-1} \,B + sec ^{-1} \,C is equal to

A

π2\frac {\pi}{2}

B

π6\frac {\pi}{6}

C

00

D

5π6\frac {5 \pi}{6}

Answer

5π6\frac {5 \pi}{6}

Explanation

Solution

Given,
(x+1)2x3+x=Ax+Bx+Cx2+1         ...(i)\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}\,\,\,\,\,\,\,\,\,...(i)
⇒(x+1)2=A(x2+1)+(Bx+c)(x)\Rightarrow (x+1)^{2}=A\left(x^{2}+1\right)+(B x+c)(x)
x2+1+2x=Ax2+A+Bx2+Cxx^{2}+1+2 x=A x^{2}+A+B x^{2}+C x
=(A+B)x2+Cx+A=(A+B) x^{2}+C x+A
On comparing the coefficient of like powers on both sides, we get
A+B=1         ...(ii)A+B=1\,\,\,\,\,\,\,\,\,...(ii)
and C=2,    A=1C=2, \,\,\,\,A=1
From E (ii), B=0B=0
Then, sin⁡−1A+tan⁡−1B+sec⁡−1C\sin ^{-1} A+\tan ^{-1} B+\sec ^{-1} C
=sin⁡−1(1)+tan⁡−1(0)+sec⁡−1(2)=\sin ^{-1}(1)+\tan ^{-1}(0)+\sec ^{-1}(2)
=sin⁡−1sin⁡π2+tan⁡−1tan⁡0+sec⁡−1sec⁡π3=\sin ^{-1} \sin \frac{\pi}{2}+\tan ^{-1} \tan 0+\sec ^{-1} \sec \frac{\pi}{3}
=π2+0+π3=5π6=\frac{\pi}{2}+0+\frac{\pi}{3}=\frac{5 \pi}{6}