If x+12x3+x=Ax+Bx+Cx2+1, then cosec–11A+cot-11B+sec-1C= (a) 0 (b) π2 - Maths - Inverse Trigonometric Functions

Question

If x + 1 2 x 3 + x = A x + B x + C x 2 + 1 ,   then  
cosec – 1 1 A + cot - 1 1 B + sec - 1 C =
(a) 0 (b) π 2 (c) 5 π 6 (d) π 6

Aarushi Mishra · Accepted Answer

x+12x3+x=Ax+Bx+Cx2+1x+12xx2+1=Ax+Bx+Cx2+1x+12=xx2+1Ax+xx2+1Bx+Cx2+1x+12=Ax2+1+xBx+CPut x=012=AA=1x+12=x2+1+xBx+Cx2+1+2x=x2+1+xBx+C2x=xBx+C2x=Bx2+CxComparing both sidesB=0C=2cosec-11A+cot-11B+sec-1C=cosec-111+cot-110+sec-12=cosec-111+cot-1∞+sec-12=π2+0+π3=5π6

If x + 1 2 x 3 + x = A x + B x + C x 2 + 1 , then cosec – 1 1 A + cot - 1 1 B + sec - 1 C = (a) 0 (b) π 2 (c) 5 π 6 (d) π 6

If $\frac{{(x + 1)}^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}, then {cosec}^{- 1} (\frac{1}{A}) + \cot^{- 1} (\frac{1}{B}) + \sec^{- 1} (C) =$
(a) 0 (b) $\frac{π}{2}$ (c) $\frac{5 π}{6}$ (d) $\frac{π}{6}$