Rationalise the denominator.

(1 - sinx)/(1 + sinx)

= (1 - sinx)/(1 + sinx) × (1 - sinx)/(1 - sinx)

= (1 - sinx)² / ((1 + sinx)(1 - sinx))

= (1 - sinx)² / (1² - sin²x)

= (1 - sinx)² / cos²x

= ((1 - sinx)/cosx)²

= (1/cosx - sinx/cosx)²

= (secx - tanx)²


So √((1 - sinx)/(1 + sinx))

= √(secx - tanx)²

= secx - tanx

(Proved)

Why must be it be acute ?

From the last part from previous section :

√((1 - sinx)/(1 + sinx))

= √(secx - tanx)²

We are square rooting a squared term here.

The result must be ≥ 0 since we are only considering the square root of real numbers which are ≥ 0 . Square rooting negative real numbers will give a complex number result.


When x is acute, 0°< x < 90°

0 < cosx < 1 (Recall ASTC , first quadrant)

So, 1/cosx > 1 (technically it can approach positive infinity as cosx → 0)

secx > 1

But when 0° < x < 90°,

0 < tanx < 1 (ASTC, first quadrant)

So secx is definitely bigger than tanx in this range/domain.

Therefore secx - tanx > 0


So in order to get a positive/non-negative result from square rooting, it is represented as secx - tanx

But if x is obtuse , 0° < x < 180°

-1 < cosx < 0 (ASTC, second quadrant)

So 1/cosx < -1 (it approaches negative infinity as cosx → 0)

secx < -1

And,

-1 < tan x < 0 (ASTC,second quadrant)

So , secx < tanx and therefore secx - tanx < 0
And tanx - secx > 0


But the square root of a real number ≥ 0 can never be negative.

So we use tanx - secx to represent it since tanx - secx > 0

If x can take any real value where sinx ≠ -1, this means that x ≠ 270° + n(360°) , where n is an integer.

i.e x ≠ 270°, 630°, 990°,...
x ≠ -90°, -450°, -810°,...


So 1 + sinx ≠ 0, the denominator is non-zero and then the expression √((1 - sinx)/(1+sinx)) will not be undefined.

We saw that for acute angles, secx - tanx is positive while for obtuse angles, tanx - secx is positive.


But what about the general case? Which includes :

0°
right angles (90°)
straight angles (180°)
reflex angles (>180° but <360°)
full/complete angles (360°)
angles beyond 360°
And their corresponding negative angles ?


Eg. If x =... -270° ,90°,450°,810°,...

sinx = 1 so 1 - sinx = 0, 1 + sinx = 2

Then √((1 - sinx)/(1+sinx)) = √(0/2) = 0

Yet, cosx is 0 for x = 90°,450°,810°...
So secx (= 1/cosx) is undefined.

And, since tanx = sinx / cosx, tanx is also undefined for these values as the denominator is also 0.

So for these values, we should not use secx - tan x or tanx - secx to represent the expression

It will just be √((1 - sinx)/(1 + sinx)) = 0


(Side note : it may be arguable that secx - tanx and tanx - secx = 0 for these values, since :

tanx = sinx/cosx and sinx = 1 for x = 90°, 450°,810,...

so tanx = 1/cosx = secx

And so secx - tanx = 0 and tanx - secx = 0

i.e the two terms cancel out)

As for the other values of x,

since we are talking about the general case, we cannot use secx - tanx as it will be negative for some values of x.

Likewise, tanx - secx will also be negative for some values of x.


We need an expression that will always be positive.

And this is solved by using the modulus sign (which gives the absolute value)


So for ii),

Either |secx - tanx| or |tanx - secx| is acceptable (they are equivalent)