Question

if y=√(1-sin4x)+1/√(1+sin4x)-1 then one of the values of y is/are

Answer 1

y=sec2x+tan2x.............................

Answer 2

$y = \frac{ \sqrt{(1 - \sin(4x)) } + 1 }{ \sqrt{ \sin(4x) + 1} - 1} \\ y = \frac{ \sin(2x) - \cos(2x) + 1 }{ \sin(2x) + \cos(2x) - 1 } \\ y = \frac{2 \sin(x) \cos(x) + 2 { \sin(x) }^{2} }{ 2\sin(x) \cos(x) - 2 { \sin(x) }^{2} } \\ y = \frac{ \cos(x) + \sin(x) }{ \cos(x) - \sin(x) } \\ y = \frac{ \sin( \frac{\pi}{4} + x ) }{ \cos( \frac{\pi}{4} + x) } \\ y = \tan( \frac{\pi}{4} + x)$

formula used

Sin(2A) = 2sinAcosA

1 - cos2A = 2sin^2A

1 - sin2A = (sin A - cosA)^2

1+ sin2A = (sinA + cosA)^2

sinA + cosA = sin(π/4 + X)/sqrt,2 = cos(π/4 -X))sqrt2