Tori Hines

2022-01-24

Why must x be acute for the identity $\sqrt{\frac{1-\mathrm{sin}x}{1+\mathrm{sin}x}}=\mathrm{sec}x-\mathrm{tan}x$ to hold true?

izumrledk

The first quadrant is ${0}^{\circ }\le x\le {90}^{\circ }$ and the fourth quadrant is ${270}^{\circ }\le x\le {360}^{\circ }$ or this is the same as $-{90}^{\circ }\le x\le {0}^{\circ }$. That is
$-{90}^{\circ }\le x\le {90}^{\circ }$
represents both the first and the fourth quadrant.

Serifluinueyk

The important point is that “identity” is true only when $-{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}^{\circ } 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 ${330}^{\circ }$ in the anticlockwise direction is the same as moving it through ${30}^{\circ }$ in the clockwise direction.

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