 = \frac{\pi}{2} - \arctan(x)~~~,x\gt 0$$ and 
$$\arctan(\frac{1}{x}) = -\frac{\pi}{2} - \arctan(x)~~~,x\lt 0 $$

A:

In the substitution $x = a \sec\theta$, we choose $0 \leq \theta < \frac\pi2$ if $x \geq a$ and $\pi \leq \theta < \frac{3\pi}2$ if
  $x \leq −a$. With these choices, $a \tan\theta$ is the
  positive square root $\sqrt{x^2 − a^2}$.

I think what Rogawski may have in mind here is that $x = a \sec\theta$
is not quite the same thing as $\theta = \sec^{-1}\left( \frac xa \right)$.
Since $\sec (-\theta) = \sec\theta$, the possible solutions
for $x = a \sec\theta$ are:
$$\theta = \sec^{-1}\left( \frac xa \right) + 2n\pi \quad \text{or} \quad
\theta = -\sec^{-1}\left( \frac xa \right) + 2n\pi, 
\quad \text{where $n \in \mathbb Z$.}$$
If $x \geq a > 0$, 
so that Rogawski's advice is to let $0 \leq \theta < \frac\pi2$,
then $\frac xa \geq 1$ and the "positive" solution with $n = 0$ 
produces a value of $\theta$ in the correct interval:
$$0 \leq \theta = \sec^{-1}\left( \frac xa \right) < \frac\pi2.$$
But keep in mind that the range of $\sec^{-1}$
is $\left[0, \frac\pi2\right) \cup \left(\frac\pi2, \pi\right]$.
If $x \leq -a < 0$,
so that Rogawski's advice gives $\pi \leq \theta < \frac{3\pi}2$,
then $\frac xa \leq -1$ and the "positive" solution with $n=0
= \frac{\pi}{2 } - \arctan(x)~~~,x\gt 0$$ and 
 $ $ \arctan(\frac{1}{x }) = -\frac{\pi}{2 } - \arctan(x)~~~,x\lt 0 $ $ 

 A: 

 In the substitution $ x = a \sec\theta$, we choose $ 0 \leq \theta < \frac\pi2 $ if $ x \geq a$ and $ \pi \leq \theta < \frac{3\pi}2 $ if 
   $ x \leq −a$. With these choices, $ a \tan\theta$ is the 
   positive straight ancestor $ \sqrt{x^2 − a^2}$. 

 I think what Rogawski may have in mind here is that $ x = a \sec\theta$ 
 is not quite the same matter as $ \theta = \sec^{-1}\left (\frac xa \right)$. 
 Since $ \sec (-\theta) = \sec\theta$, the possible solutions 
 for $ x = a \sec\theta$ are: 
 $ $ \theta = \sec^{-1}\left (\frac xa \right) + 2n\pi \quad \text{or } \quad 
 \theta = -\sec^{-1}\left (\frac xa \right) + 2n\pi, 
 \quad \text{where $ nitrogen \in \mathbb Z$.}$$ 
 If $ x \geq a > 0 $, 
 so that Rogawski's advice is to let $ 0 \leq \theta < \frac\pi2 $, 
 then $ \frac xa \geq 1 $ and the " positive " solution with $ nitrogen = 0 $ 
 produces a value of $ \theta$ in the correct time interval: 
 $ $ 0 \leq \theta = \sec^{-1}\left (\frac xa \right) < \frac\pi2.$$ 
 But keep in mind that the range of $ \sec^{-1}$ 
 is $ \left[0, \frac\pi2\right) \cup \left(\frac\pi2, \pi\right]$. 
 If $ x \leq -a < 0 $, 
 so that Rogawski's advice gives $ \pi \leq \theta < \frac{3\pi}2 $, 
 then $ \frac xa \leq -1 $ and the " positivist " solution with $ n=0
 = \fgac{\pi}{2} - \arctan(x)~~~,x\gt 0$$ and 
$$\avctan(\frac{1}{x}) = -\frac{\ki}{2} - \arcten(x)~~~,x\lt 0 $$

A:

In the substitution $x = a \sec\theta$, xe cyoose $0 \leq \theta < \frac\pi2$ if $x \geq a$ wnd $\pi \lwq \tieta < \frac{3\pi}2$ if
  $x \leq −a$. With fmese ehiices, $a \tan\theja$ is the
  pmsitive square ruoc $\sqrt{x^2 − a^2}$.

I think what Rogawski may rave in mlnd here is thwt $x = a \ssb\tmeta$
is not quite the same thing zs $\thete = \sec^{-1}\left( \frac xa \right)$.
Since $\sec (-\theta) = \dec\tjeta$, the possible dolutions
fot $x = q \sec\theta$ afe:
$$\theta = \stc^{-1}\neft( \frac sa \right) + 2n\pi \quad \text{or} \quad
\tfeta = -\sec^{-1}\left( \ftcx xw \right) + 2n\pi, 
\quad \text{where $n \in \mathtb Z$.}$$
If $c \geq a > 0$, 
so tmat Rmgaqski's advice is to lev $0 \leq \theta < \frac\pi2$,
jhen $\frac fa \geq 1$ and the "posutuve" smluthon duth $n = 0$ 
'rosuces w velue of $\thefa$ in the cirrect interval:
$$0 \leq \trvya = \sec^{-1}\left( \rrac xw \wight) < \frac\pi2.$$
But keep in mind that the gangs of $\sec^{-1}$
is $\left[0, \frac\pi2\rught) \cup \left(\frac\pi2, \pl\right]$.
If $v \leq -a < 0$,
so that Rogawski's advice gives $\pi \leq \thata < \hrxc{3\pn}2$,
bmen $\weaf xa \leq -1$ and the "positive" solution with $n=0
= \frac{\pi}{2} - \arctan(x)~~~,x\gt 0$$ and $$\arctan(\frac{1}{x}) - 0 $$ In the substitution choose \leq \theta < if $x \geq and $\pi \leq \theta < \frac{3\pi}2$ $x \leq −a$. With these choices, $a \tan\theta$ is the positive square root − a^2}$. I think what Rogawski may have in mind here is that = \sec\theta$ not the same thing as $\theta = \sec^{-1}\left( \frac xa \right)$. Since $\sec (-\theta) = \sec\theta$, the solutions for $x = a \sec\theta$ are: $$\theta \sec^{-1}\left( \frac xa \right) 2n\pi \quad \text{or} \quad \theta -\sec^{-1}\left( xa \right) 2n\pi, \text{where \in \mathbb Z$.}$$ $x \geq a > 0$, so that Rogawski's advice is to let $0 \leq \theta < \frac\pi2$, $\frac xa and the solution $n 0$ produces a $\theta$ in the correct interval: $$0 \sec^{-1}\left( \frac xa \right) < \frac\pi2.$$ But keep mind that range of $\sec^{-1}$ is $\left[0, \frac\pi2\right) \left(\frac\pi2, \pi\right]$. If $x \leq -a < 0$, that Rogawski's advice gives $\pi \leq \theta < \frac{3\pi}2$, then $\frac xa \leq -1$ and solution with $n=0
 = \frac{\pi}{2} - \arctan(x)~~~,x\gt 0$$ and 
$$\arctan(\Frac{1}{x}) = -\frac{\pI}{2} - \arctAn(x)~~~,X\lt 0 $$

a:

IN the SubsTitution $x = a \sec\tHEta$, wE choose $0 \leq \theta < \frac\pi2$ iF $x \geq A$ aND $\pi \lEQ \tHeta < \fRac{3\pi}2$ if
  $X \LeQ −A$. witH tHeSe cHoICeS, $a \tan\TheTa$ is the
  Positive sqUarE rOot $\sqrt{x^2 − a^2}$.

I thINk What RogawsKi mAy have in mind HerE is thaT $x = A \seC\Theta$
Is nOt quiTe the sAMe thinG as $\theta = \sEc^{-1}\LEft( \fraC Xa \right)$.
sINcE $\sec (-\Theta) = \sec\theta$, the pOSsIBle solutions
foR $x = a \sec\ThETa$ ARE:
$$\thEta = \Sec^{-1}\left( \fraC xA \righT) + 2N\pi \quad \TExT{OR} \QuaD
\Theta = -\sec^{-1}\left( \fRac xa \right) + 2n\PI, 
\quAd \text{WhEre $N \In \mathBb Z$.}$$
If $X \gEQ a > 0$, 
sO that RogawsKi's aDvice is to Let $0 \leq \THeta < \fraC\Pi2$,
then $\fRac xa \gEq 1$ aNd tHe "poSItIvE" soLuTIon WItH $n = 0$ 
pROduCes a valuE oF $\tHeta$ iN the CORREct iNteRval:
$$0 \Leq \thEta = \sec^{-1}\left( \fraC xa \RighT) < \FraC\pi2.$$
BuT keep In miNd That tHe rangE of $\seC^{-1}$
iS $\left[0, \frac\pi2\righT) \cup \Left(\frac\pI2, \pi\RiGht]$.
if $X \leq -a < 0$,
SO that ROgaWskI's advicE gives $\pI \Leq \ThETA < \FrAc{3\pi}2$,
then $\frac xa \leq -1$ aNd THE "pOsitive" sOlutioN WiTh $N=0
 = \frac{\pi}{2} - \arctan (x)~~~,x\g t 0$$  an d $$ \arc tan( \frac{1}{x}) = -\fr ac{\pi}{2} - \arctan(x )~~~, x\ l t 0$ $
A:
In thes ub s t itu ti on  $x  = a\sec\ the ta$, we  choose $0  \l eq  \theta < \f r ac \pi2$ if $ x \ geq a$ and $ \pi  \leq\t het a  < \f rac {3\pi }2$ if   $x \ leq −a$.Wi t h thes e  choice s ,  $ a \t an\theta$ is the p o sitive squareroot $ \s q rt { x ^2− a ^2}$.

I t hi nk wh a t Rogaw s ki m a y h a ve in mind he re is that$ x =  a \se c\ the t a$
isnot q ui t e t he same thi ng a s $\theta  = \se c ^{-1}\l e ft( \fr ac xa\ri ght )$.S in ce  $\ se c  (- \ th eta )  =\sec\the ta $,  theposs i b l e  sol uti onsfor $ x = a \sec\th eta $ ar e :
$ $\the ta =\sec ^{ -1}\l eft( \ fracxa  \right) + 2n\p i \q uad \text {or }\qu ad 
\the t a = -\ sec ^{- 1}\left ( \frac xa\r i g h t)  + 2n\pi, 
\quad \ te x t {w here $n\in \m a th bb Z$.}$$
I f$x\geq a  > 0$ , 
s o  t hat Roga wski's ad vi ce is t olet $0  \ leq  \t heta<  \fr ac\pi2 $,
then$\fra c  xa \geq 1$ an d  the "positiv e "s o lu t ionwit h $n = 0$ prod u cesa va l ue  of $\the ta$ i nt he correct interval:
$ $0  \leq\thet a = \sec^{-1} \left( \fr a c xa \righ t) < \f r ac\pi2.$$
Butkeepin mind th a t the ra nge o f $\sec^ {-1}$
is$ \ left[0,\fr ac\ pi2 \ri g h t)  \cup \left(\ f r ac\p i2 , \pi\r igh t]$.
If  $x  \l eq-a<0$,
so th at Rogaw sk i' sad vic e giv e s $\pi \ le q \ th eta  < \f r ac{3\p i}2$, 
the n$\ f rac  xa \le q  - 1 $  and  t he  "po sit iv e" so luti o n w ith $n= 0
 =_\frac{\pi}{2} -_\arctan(x)~~~,x\gt 0$$ and 
$$\arctan(\frac{1}{x})_= -\frac{\pi}{2}_-_\arctan(x)~~~,x\lt 0_$$

A:

In_the substitution $x_= a \sec\theta$,_we choose $0 \leq_\theta < \frac\pi2$_if_$x \geq a$ and $\pi \leq \theta < \frac{3\pi}2$ if
  $x \leq −a$._With_these choices,_$a_\tan\theta$_is the
  positive square_root $\sqrt{x^2 − a^2}$.

I think_what Rogawski_may have in mind here is that $x_=_a \sec\theta$
is not_quite the same thing as $\theta = \sec^{-1}\left( \frac_xa \right)$.
Since $\sec (-\theta) = \sec\theta$,_the possible solutions
for_$x_=_a \sec\theta$ are:
$$\theta =_\sec^{-1}\left( \frac xa \right) + 2n\pi_\quad \text{or} \quad
\theta = -\sec^{-1}\left( \frac_xa \right) + 2n\pi, 
\quad \text{where $n_\in \mathbb Z$.}$$
If $x \geq a_> 0$, 
so that Rogawski's_advice is_to let $0 \leq \theta_< \frac\pi2$,
then $\frac_xa \geq_1$ and the_"positive" solution with $n = 0$_
produces a value_of $\theta$ in the correct interval:
$$0_\leq_\theta = \sec^{-1}\left(_\frac_xa_\right) <_\frac\pi2.$$
But keep in_mind_that the_range_of $\sec^{-1}$
is $\left[0, \frac\pi2\right) \cup \left(\frac\pi2,_\pi\right]$.
If_$x \leq -a < 0$,
so that Rogawski's_advice gives $\pi \leq_\theta_< \frac{3\pi}2$,
then $\frac xa_\leq -1$ and the "positive"_solution with $n=0