Cfrgtkky

I have the following question:

Prove that: $$ cot^{-1}Biggl(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}Biggl) = frac x2, x in biggl(0, frac pi4biggl) $$

The solution:

My question is about the second step (i.e., highlighted). We know that it has came from the following resolution: $$ sqrt{1pm sin x} = sqrt{sin^2frac x 2 + cos^2frac x 2 pm2sinfrac x 2cosfrac x 2} = pm biggl( cosfrac x 2 pm sinfrac x 2 biggl) $$

I've included the $pm$ symbol in accordance with the standard definition of square root. The above solution uses the positive resolution (i.e., $ cosfrac x 2 - sinfrac x 2 $) for $ sqrt{1- sin x} $. But, if I use the negative one, then the result changes. The sixth step changes to: $$ cot^{-1}biggl(tanfrac x 2biggl) = cot^{-1}Biggl(cotleft(frac pi 2 - frac x 2right)Biggl) $$

which yield the result: $$ frac pi 2 - frac x 2 $$

Mathematically, this result is different from that provided in the RHS of question.

Is the question statement wrong or I've been hacked up?

Rationalize the numerator to find

$$f(x)=dfrac{1+|cos x|}{sin x}$$

Now if $cos xge0,$ $$f(x)=cotdfrac x2$$

$cot^{-1}f(x)=?$

Else $f(x)=dfrac{1-cos x}{sin x}=tandfrac x2$

Now $cot^{-1}f(x)=dfracpi2-tan^{-1}f(x)=?$

As StubbornAtom mentioned in comment, for $x in biggl(0, frac pi4biggl)$, the square root is:
$$sqrt{1-sin x}=cos frac x2-sin frac x2,$$
because:
$$cos frac x2-sin frac x2>0.$$
If, for example, $x in biggl(frac pi2, pibiggl)$, the square root would be:
$$sqrt{1-sin x}=sinfrac x2-cos frac x2,$$
because:
$$sin frac x2-cos frac x2>0.$$

Addendum. Note that $pm$ is placed to make the number positive and it does not imply one gets positive and negative values for the square root. Here is another way to write the same equality:
$$sqrt{1-sin x}=left|cos frac x2-sin frac x2right|=begin{cases}+left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2ge 0 \ -left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2< 0 \ end{cases}.$$

Note that for $xin (0,{pi over 4})$ we have $$sqrt{1-{sqrt 2over 2}}<sqrt{1-sin x}<1\sqrt{1-{sqrt 2over 2}}=cos {pi over 8}-sin {pi over 8}<cos {xover 2}-sin{xover 2}<1\-1<sin {x over 2}-cos {x over 2}<-sqrt{1-{sqrt 2over 2}}$$comparing the range for all $xin (0,{pi over 4})$ we obtain$$sqrt{1-sin x}=cos {xover 2}-sin{xover 2}$$and$$sqrt{1-sin x}ne sin {xover 2}-cos{xover 2}$$

We see that
$$frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
$$= sqrt{(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})^2}$$
$$= sqrt{frac{1+sin x +1-sin x + 2cdotsqrt{1-sin^2 x}}{1+sin x +1-sin x -2cdotsqrt{1-sin^2 x}}}$$
$$= sqrt{frac{1+cos x}{1-cos x}}$$
$$= sqrt{frac{2cdotcos^2frac{x}{2}}{2cdotsin^2frac{x}{2}}}$$
$$= cot frac{x}{2}$$

And hence,
$$cot^{-1} (frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})$$
$$= cot^{-1} (cot frac{x}{2})$$
$$= frac{x}{2}$$

Note: as $xin [0,frac{pi}{4}]$, $sqrt{1-sin x} le sqrt{1+sin x}$.
And hence
$$0lefrac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$