Remember that

$cos theta=dfrac{text{adjacent;side}}{text{hypotenuse}}$

$sin theta=dfrac{text{opposite;side}}{text{hypotenuse}}$

$tan theta=dfrac{text{opposite;side}}{text{adjacent side}}$

$sectheta=dfrac{1}{costheta}=dfrac{text{hypotenuse}}{text{adjacent side}}$

$csctheta=dfrac{1}{sin theta}=dfrac{text{hypotenuse}}{text{opposite side}}$

$cot theta=dfrac{1}{tan theta}=dfrac{text{adjacent side}}{text{opposite side}}$

Remember the Pythagorean formula $c^2=a^2+b^2$ where $c$ is the hypotenuse and $a$, $b$ are legs.

a)
hypotenuse = $sqrt{8^2+13^2}=sqrt{233}$

adjacent side = 8

opposite side = 13

$sec theta=dfrac{text{hypotenuse}}{text{adjacent side}}=dfrac{sqrt{233}}{8}$

$csc theta=dfrac{text{hypotenuse}}{text{opposite side}}=dfrac{sqrt{233}}{13}$

$cot theta=dfrac{text{adjacent side}}{text{opposite side}}=dfrac{8}{13}$

We can calculate $theta$ using any trigonometric ratio,

$tan theta=dfrac{13}{8}$

$theta=tan^{-1}left(dfrac{13}{8}right)=58^circ$

b) hypotenuse = $sqrt{12^2+35^2}=sqrt{1369}=37$

opposite side = 12

adjacent side = 35

$sec theta=dfrac{37}{35}$

$csctheta=dfrac{37}{12}$

$cot theta=dfrac{35}{12}$

$$
theta=tan^{-1}left(dfrac{12}{35}right)=19^circ
$$

c) adjacent side $=sqrt{39^2-23^2}=sqrt{992}=sqrt{16cdot 62}=4sqrt{62}$

opposite side = $23$

hypotenuse = 39

$sec theta=dfrac{39}{4sqrt{62}}$

$csc theta=dfrac{39}{23}$

$cot theta=dfrac{4sqrt{62}}{23}$

$$
theta=sin^{-1}left(dfrac{23}{39}right)=36^circ
$$

a) $csctheta=dfrac{sqrt{233}}{13}$ , $sectheta=dfrac{sqrt{233}}{8}$ , $cottheta=dfrac{8}{13}$ , $theta=58^circ$

b) $csctheta=dfrac{37}{12}$ , $sectheta=dfrac{37}{35}$ , $cottheta=dfrac{35}{12}$ , $theta=19^circ$

c) $csctheta=dfrac{29}{23}$ , $sectheta=dfrac{39}{4sqrt{62}}$ , $cottheta=dfrac{4sqrt{62}}{23}$ , $theta=36^circ$

The coordinates of terminal points $(cos theta, sin theta)$ for the special angle $theta$ in a unit circle are the following:

$thetaimplies (cos theta,sin theta)$

$theta=30^circimplies left(dfrac{sqrt{3}}{2},dfrac{1}{2}right)$

$theta=45^circimplies left(dfrac{sqrt{2}}{2}, dfrac{sqrt{2}}{2}right)$

$theta=60^circimplies left(dfrac{1}{2},dfrac{sqrt{3}}{2}right)$

$tan theta=dfrac{sin theta}{costheta}$

a) $left(dfrac{sqrt{2}}{2}right)left(dfrac{sqrt{2}}{2}right)+left(dfrac{1}{2}right)left( dfrac{1}{2}right)=dfrac{1}{2}+dfrac{1}{4}=dfrac{3}{4}$

b) $left(1-dfrac{sqrt{2}/2}{sqrt{2}/2}right)left(dfrac{1}{2}right)left(dfrac{sqrt{3}}{2}right)left(dfrac{sqrt{3}/2}{1/2}right)=0$

c) $dfrac{1/2}{sqrt{3}/2}+2left(dfrac{sqrt{2}}{2}right)left(dfrac{1}{2}right)=dfrac{1}{sqrt{3}}+dfrac{sqrt{2}}{2}$

$$
=dfrac{2+sqrt{2}sqrt{3}}{2sqrt{3}}cdot dfrac{sqrt{3}}{sqrt{3}}=dfrac{2sqrt{3}+3sqrt{2}}{6}
$$

a) $dfrac{3}{4}$

b) $0$

c) $dfrac{2sqrt{3}+3sqrt{2}}{6}$

begin{table}[]
defarraystretch{1.4}%
begin{tabular}{|l|l|l|l|l|}
hline
& QI & QII & QIII & QIV\ hline
interval & $0<theta<90^circ$ & $90^circ< theta<180^circ$ & $180^circ <theta<270^circ$ & $270^circ <theta<360^circ$ \ hline
reference angle $beta$ & $theta$ & $180^circ-theta$ & $180^circ+theta$ & $360^circ-theta$ \ hline
sine & $+$ & $+$ & $-$ & $-$ \ hline
cosine & $+$ & $-$ & $-$ & $+$ \ hline
tangent & $+$ & $-$ & $+$ & $-$ \ hline
end{tabular}

end{table}

a) $tan 18^circ=0.3249$

(i) $18^circ$ is in QI where $tan theta>0$

(ii) $theta=18^circ$, $beta=18^circ$

Another angle that has equivalent ratio is in QIII with $beta=18^circ$

$$
theta=180^circ+18^circ=198^circ
$$

b) $sin 205^circ=-0.4226$

(i) $205^circ$ is in QIII where $sin theta<0$

(ii) $beta=theta-180^circ=205^circ-180^circ=25^circ$

Another angle that has equivalent ratio is in QIV with $beta=25^circ$

principal angle: $theta=360-beta=360-25=335^circ$

c)

(i) $cos(-55^circ)=0.5736$

It goes clockwise so it will be in QIV where $cos theta>0$

(ii) $beta=55^circ$

Another angle with equivalent ratio is in QI with $beta=55^circ$

principal angle: $55^circ$

a) $+$ ; $tan 18^circ=0.3249$ ; $theta=198^circ$, $beta=18^circ$

b) $-$ ; $sin 205^circ=-0.4226$ ; $theta=355^circ$ , $beta=25^circ$

c) $+$ ; $cos(-55^circ)=0.5736$ ; $theta=55^circ$ , $beta=55^circ$

If $P(x,y)$ is the terminal point of $theta$

$r=sqrt{x^2+y^2}$

$cos theta=dfrac{x}{r}$

$sin theta=dfrac{y}{r}$

$tan theta=dfrac{y}{x}$

a) $r=sqrt{(-2)^2+5^2}=sqrt{29}$

$cos theta=dfrac{-2}{sqrt{29}}=-dfrac{2sqrt{29}}{29}$

$sin theta=dfrac{5}{sqrt{29}}=dfrac{5sqrt{29}}{29}$

$$
tan theta=dfrac{5}{-2}=-dfrac{2}{5}
$$

b) $r=sqrt{(3)^2+(-3)^2}=sqrt{18}=sqrt{9cdot 2}=3sqrt{2}$

$cos theta=dfrac{3}{3sqrt{2}}=dfrac{1}{sqrt{2}}=dfrac{sqrt{2}}{2}$

$sin theta=dfrac{-3}{3sqrt{2}}=-dfrac{1}{sqrt{2}}=-dfrac{sqrt{2}}{2}$

$$
tan theta=dfrac{-3}{3}=-1
$$

c) $r=sqrt{(-4)^2+(-5)^2}=sqrt{41}$

$cos theta=dfrac{-4}{sqrt{41}}=-dfrac{4sqrt{41}}{41}$

$sin theta=dfrac{-5}{sqrt{41}}=-dfrac{5sqrt{41}}{41}$

$$
tan theta=dfrac{-5}{-4}=dfrac{5}{4}
$$

a) $sin theta=dfrac{5sqrt{29}}{29}$ , $cos theta=dfrac{-2sqrt{29}}{29}$ , $tan theta=-dfrac{5}{2}$

b) $sin theta=-dfrac{sqrt{2}}{2}$ , $costheta=dfrac{sqrt{2}}{2}$ , $tan theta=-1$

c) $sin theta=-dfrac{5sqrt{41}}{41}$ , $costheta=-dfrac{4sqrt{41}}{41}$ , $tantheta=dfrac{5}{4}$

a) $cos theta$ is negative in QII and QIII

$cos phi = dfrac{x}{r}=-dfrac{7}{sqrt{53}}$

Using the identity $sin^2phi +cos^2phi =1$

$sin phi=pmsqrt{1-cos^2phi}$

$$
sin phi =pmsqrt{1-left(dfrac{-7}{sqrt{53}}right)^2}=pmsqrt{dfrac{4}{53}}=pm dfrac{2}{sqrt{53}}
$$

There are two cases.

b) If $phi$ is in QII

$sin phi = dfrac{2}{sqrt{53}}$

$tan phi = dfrac{sinphi}{cosphi}=dfrac{2/sqrt{53}}{-7/sqrt{53}}=-dfrac{2}{7}$

$sec phi = dfrac{1}{costheta}=-dfrac{sqrt{53}}{7}$

$cscphi=dfrac{1}{sin theta}=dfrac{sqrt{53}}{2}$

$cotphi=dfrac{1}{tanphi}=-dfrac{7}{2}$

If $phi$ is in QIII,

$sin phi = -dfrac{2}{sqrt{53}}$

$tan phi = dfrac{2}{7}$

$sec phi=-dfrac{sqrt{53}}{7}$

$csc phi =dfrac{sqrt{53}}{2}$

$$
cot theta=dfrac{7}{2}
$$

We shall evaluate $phi$, to do that, we must find the reference angle $beta$

$phi=cos^{-1}left(left|dfrac{-7}{sqrt{53}}right|right)=16^circ$

In QII: $phi = 180^circ-beta=180^circ-16^circ=164^circ$

In QIII: $phi=180^circ+beta=180^circ+16^circ=196^circ$

a) Quadrant 2 or 3

b) Quadrant 2: $sin phi = dfrac{2}{sqrt{53}}$ , $tan phi=-dfrac{2}{7}$ , $cscphi=dfrac{sqrt{53}}{2}$ ,

$secphi=-dfrac{sqrt{53}}{7}$ , $cotphi=-dfrac{7}{2}$

Quadrant 3: $sinphi=-dfrac{2}{sqrt{53}}$ , $tan phi=dfrac{2}{7}$ , $cscphi=-dfrac{sqrt{53}}{2}$ , $secphi=-dfrac{sqrt{53}}{7}$ , $cotphi=dfrac{7}{2}$

c) $phi =164^circ$ in quadrant 2 and $phi=196^circ$ in quadrant 3

begin{table}[]
defarraystretch{2.5}%
begin{tabular}{|l|l|}
hline
$bold{Reciprocal; Identities}$ & $sec theta=dfrac{1}{costheta}$ \ cline{2-2}
& $csc theta=dfrac{1}{sin theta}$ \ cline{2-2}
& $cot theta=dfrac{1}{tan theta}$ \ hline
$bold{Quotient; Identities}$ & $tan theta=dfrac{sintheta}{costheta}$ \ cline{2-2}
& $cottheta=dfrac{costheta}{sin theta}$ \ hline
$bold{Pythagorean; Identities}$ & $sin^2theta+cos^2theta=1$ \ cline{2-2}
& $1+tan^2theta=sec^2theta$ \ cline{2-2}
& $1+cot^2theta=csc^2theta$ \ hline
end{tabular}
end{table}

L.S. = $cos beta cot beta=cos beta cdot dfrac{cosbeta}{sin beta}$

$=dfrac{cos^2beta}{sin beta}=dfrac{1-sin^2beta}{sin beta}=dfrac{1}{sin beta}-sin beta$ = R.S.

restriction: $sin betane 0$

begin{table}[]
defarraystretch{2.5}%
begin{tabular}{|l|l|}
hline
$bold{Reciprocal; Identities}$ & $sec theta=dfrac{1}{costheta}$ \ cline{2-2}
& $csc theta=dfrac{1}{sin theta}$ \ cline{2-2}
& $cot theta=dfrac{1}{tan theta}$ \ hline
$bold{Quotient; Identities}$ & $tan theta=dfrac{sintheta}{costheta}$ \ cline{2-2}
& $cottheta=dfrac{costheta}{sin theta}$ \ hline
$bold{Pythagorean; Identities}$ & $sin^2theta+cos^2theta=1$ \ cline{2-2}
& $1+tan^2theta=sec^2theta$ \ cline{2-2}
& $1+cot^2theta=csc^2theta$ \ hline
end{tabular}
end{table}

a) L.S. = $tan alpha cos alpha$

$=left(dfrac{sin alpha}{cos alpha}right)(cos alpha)$

$=sin alpha$ = R.S.

b) R.S. $= sin phi secphi$

$= sin phi left( dfrac{1}{cos phi}right)$

$=dfrac{sinphi}{cosphi}cdot dfrac{1/sin phi}{1/sinphi}$

$=dfrac{1}{(cosphi/sinphi)}$

$=dfrac{1}{cotphi}$ = L.S

$$
cosphineq 0implies phineq 90^circtext{ or } 270^circ
$$

c) R.S. = $dfrac{sin x cos x}{cot x}$

$$
=dfrac{sin x cos x}{cos x/sin x}
$$

$$
=dfrac{sin^2 xcdot cos x}{cos x}
$$

$= sin^2 x$

$=1-cos^2x$ = L.S.

$$
sin xneq 0implies xneq 0^circtext{ or }180^circ
$$

d) L.S. = $sectheta cos theta+secthetasin theta$

$=sectheta (costheta+sin theta )$

$=dfrac{1}{costheta} cdot (costheta+sin theta)$

$=dfrac{costheta}{costheta}+dfrac{sin theta}{costheta}$

$=1+tantheta$ = R.S.

$$
costhetaneq 0implies thetaneq 90^circtext{ or }270^circ
$$

begin{table}[]
begin{tabular}{|l|l|l|}
hline
$angle A < 90^circ$ & $a<h$ & no triangle exists \ cline{2-3}
& $a=h$ & one right triangle \ cline{2-3}
& $bsin A<b<a$ & one triangle \ cline{2-3}
& $bsin A<a90^circ$ & $ab$ & one triangle \ hline
end{tabular}
end{table}