Why must $x$ be acute for the identity $\sqrt{\frac{1- \sin x}{1+ \sin x}}=\sec x- \tan x$ to hold true? The question first requires me to prove the identity $$\sqrt{\frac{1- \sin x}{1+ \sin x}}=\sec x- \tan x, -90^\circ < x < 90^\circ$$ I am able to prove this. The second part says “Explain why $x$ must be acute for the identity to be true”. I don’t see why $x$ must be acute for the identity to hold true. Wouldn’t it suffice for it to lie in either the 1st or 4th quadrant? Eg $x=330^\circ$. 
 A: The first quadrant is $0^\circ \leq x \leq 90^\circ$ and the fourth quadrant is $270^\circ \leq x \leq 360^\circ$ or this is the same as $-90^\circ \leq x \leq 0^\circ$. That is
$$-90^\circ \leq x\leq 90^\circ$$
represents both the first and the fourth quadrant.
A: For $1+\sin x\ne0$
$$f^2(x)=\dfrac{1-\sin x}{1+\sin x}=\dfrac{(1-\sin x)^2}{\cos^2x}$$
$$f(x)=\dfrac{|1-\sin x|}{|\cos x|}=\dfrac{1-\sin x}{|\cos x|}$$ as $1-\sin x\ge0$
Now $|\cos x|=+\cos x$ iff $\cos x\ge0$ i.e., if $x$ lies in the first or in the fourth quadrant
Else $|\cos x|=-\cos x$
A: The important point is that “identity” is true only when $-90^\circ < x < 90^\circ$.
In another word, it is not true when the terminal arm falls in QII and QIII.
Note that the angle lies in $-90^0 < x < 0^0$ is also an acute angle, because the minus sign only refers to moving the terminal arm in the clockwise direction. The resultant angle is still acute.
Added: To move the terminal arm through $330^0$ in the anticlockwise direction is the same as moving it through $30^0$ in the clockwise direction. 
