If x2+2xx2-4=Ax+Bx-2+Cx+2, then sin-1BC=

x2+2xx2-4=Ax+Bx-2+Cx+2 x2+2=A(x2-4)+Bx(x+2)+Cx(x-2) x=2⇒6=8B⇒B=34 x=-2⇒6=8C⇒C=34 sin-1BC=sin-1(1)=π2

If x2+2xx2-4=Ax+Bx-2+Cx+2, then sin-1BC=

x2+2xx2-4=Ax+Bx-2+Cx+2 x2+2=A(x2-4)+Bx(x+2)+Cx(x-2) x=2⇒6=8B⇒B=34 x=-2⇒6=8C⇒C=34 sin-1BC=sin-1(1)=π2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$If \frac{x^{2} + 2}{x (x^{2} - 4)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}, then \sin^{- 1} (\frac{B}{C}) =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{π}{2}$

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{x^{2} + 2}{x (x^{2} - 4)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2} x^{2} + 2 = A (x^{2} - 4) + B x (x + 2) + C x (x - 2) x = 2 \Rightarrow 6 = 8 B \Rightarrow B = \frac{3}{4} x = - 2 \Rightarrow 6 = 8 C \Rightarrow C = \frac{3}{4} \sin^{- 1} (\frac{B}{C}) = \sin^{- 1} (1) = \frac{π}{2}$