If `y = tan^(-1)(u/sqrt(1-u^2))` and `x = sec^(-1)(1/(2u^2-1))`, ` u in (0,1/sqrt2)uu(1/sqrt2,1)`, prove that `2dy/dx+ 1 = 0`.

Options

A. y
B. xy
C. 0
D. 1

Answer

C

Explanation

`y=tan^(-1).(u)/(sqrt(1-u^(2)))`
`rArr" "y=sin^(-1)u`
Also, `x=sec^(-1).(1)/(2u^(2)-1)`
`rArr" "x=cos^(-1)(2u^(2)-1)`
`y=sin^(-1)u`
`therefore" "(dy)/(dx)=(1)/(sqrt(1-u^(2)))`
`=cos^(-1)(2u^(2)-1)`
Also, `(dx)/(du)=(-1)/(sqrt(1-(2u^(2)-1)^(2)))(4u)`
`=-(4u)/(sqrt(1-4u^(4)-1+4u^(2)))`
`-(4u)/(sqrt(4u^(2)(1-u^(2))))=-(2)/(sqrt(1-u^(2)))`
`therefore" "(dy)/(dx)=-(1)/(2)`
`rArr" "2((dy)/(dx))+1=0`