If `y=(sqrt(1-sin4x)+1)/(sqrt(1+sin 4x)-1)`, then y can be

Options

A. cot x
B. `- tan x`
C. `-cot((pi)/(4)+x)`
D. `tan((pi)/(4)+x)`

Answer

A, B, C, D

Explanation

Since `1 pm sin theta = (cos.(theta)/(2)pm sin.(theta)/(2))^(2)`
`therefore y=(pm(cos 2x - sin 2x)+1)/(pm (cos 2x+sin 2x)-1)`
Taking '+' sign in Nr. And Dr. we get
`y=(1+cos2x -sin 2x)/(sin 2x-(1-cos 2x))`
`=(cos^(2)x-2sin x cos x)/(2 sin x cos x -2 sin^(2)x)=(cos x)/(sin x)=cot x`
Taking '-' sign in Nr. and Dr. we get
`y=(sin 2x + 1-cos 2x)/(-sin 2x -(1+cos 2x))`
`=(2 sin x cos x + 2 sin^(2)x)/(-2sin x cos x-2cos^(2)x)`
`=-tan x` Taking '+' sign in Nr. and '-' sign in Dr. we get
`y=(1+cos 2x-sin 2x)/(-sin 2x-(1+cos 2x))`
`=(2 cos^(2)x-2 sin x cos x)/(-2sin x cos x -2 cos^(2)x)`
`=(sin x - cos x)/(sin x + cos x)`
`= tan (x=(pi)/(4))`
`=-cot((pi)/(2)+x -(pi)/(4))`
`=-cot((pi)/(4)+ x)`
Taking '-' sign in Nr. and '+' sign in Dr. we get
`y=(sin 2x + 1-cos 2x)/(sin 2x-(1-cos 2x))`
`=(2 sin x cos x + 2 sin^(2)x)/(2 sin x cos x-2 sin^(2)x)`
`=(cos x + sin x)/(cos x - sin x)`
`=(1+tan x)/(1-tan x)`
`=tan((pi)/(4)+ x)`