# If y = sqrt (sec x -tan x / tan x + sec x) prove that dy/dx = sec x (tanx +secx)

Arun
25750 Points
6 years ago
Dear Aman

y = sqrt ( (sec x - tan x )/(tan x + sec x))
y' = [ 1/2 sqrt (secx - tanx /tan x + sec x) ]*[ksec x tanx - sec²x) (tan x + secx) - (sec x tanx + sec²x)( secx - tanx) ]/ (tanx + secx)²
When you simplify it, you will get the required result.
Shashank S
13 Points
5 years ago
y=√[(secx+tanx)/(secx-tanx)]
y=√(secx+tanx)(secx+tanx)/(secx-tanx)(secx+tanx)
y=√[secx+tanx]^2/[sec^2 x-tan^2 x]
Wkt sec^2 x-tan^2 x = 1 { from the  Identity }
y=(secx+tanx)
dy/dx={ d/dx(secx)+d/dx(tanx) }
dy/dx=secx.tanx+sec^2 x
dy/dx=secx[secx+Tanx] Hence proved.

Ritu kumari
13 Points
3 years ago
y' = {√(sec x + tan x)/2√(sec x - tanx)} * [(sec x +   tan x) (sec x * tan x - sec²x) - (sec x - tan x) (sec x * tan x + sec²x)] / (sec x + tan x)²
= {√(sec x + tan x)/2(sec x - tan x)}*[sec x(sec x + tan x)(tan x - sec x) - sec x (sec x - tan x)(sec x + tan x)]/(sec x + tan x)²
= [(sec x) (sec x + tan x) ( tan x - sec x - sec x +   tan x)]/2√(sec x - tan x)(sec x + tan x)³/²
=[2 (sec x) (tan x - sec x)]/[2√(sec²x - tan²x)]
=sec x ( tan x - sec x)
Hence proved
the question is not right. The sign will be negative.