Difficult trigo problem...for me?

Question

Prove that [ (1 - sin x) / (1 + sin x) ]^1/2 = sec x - tan x when x is acute.
(Please start from LHS and without altering the equation on RHS)

Why is the restriction on x necessary?

Thanks in advance!

Answer 1

[ (1 - sin x) / (1 + sin x) ]^1/2 = sec x - tan x

[ (1 - sin x)(1 -sin x) / (1 + sin x) (1 - sin x) ]^1/2 = sec x - tan x

[ (1 - sin x)^2 / (1 + sin x) (1 - sin x) ]^1/2
[ (1 - sin x)^2 / (1 - sin^2 x) ]^1/2
[ (1 - sin x)^2 / (cos^2 x) ]^1/2

[ (1 - sin x) / (cos x) ]
1/cosx - sin x/cosx
sec x -tanx

one limitation is cos x cannot be zero
x cannot be zero

Answer 2

Left Hand Side = √[(1 - sin x) / (1 + sin x)]

= √{(1 - sin x)² / [(1 + sin x)(1 - sinx)]}

= √{(1 - sin x)² / (1 - sin²x)}

= √{(1 - sin x)² / cos²x} = (1 - sinx) / cosx

= secx - tanx = Right Hand Side

The square root is positive and (secx - tanx) is positive when x is acute but not when it is obtuse. So x must be acute.