Exercise 1

Step 1

1 of 7

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}

Step 2

2 of 7

Here, we need to express each ratio in terms of $x$, $y$ and $r$

Remember that for a unit circle, $r=1$

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

$sintheta=dfrac{y}{r}=y$

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

The restrictions on $theta$ are its values that could make a denominator equal to zero.

Step 3

3 of 7

a)
L.S: $cot theta=dfrac{1}{tan theta}=dfrac{1}{y/x}=dfrac{x}{y}$

R.S: $dfrac{costheta}{sin theta}=dfrac{x}{y}$

Restrictions: $sin thetaneq 0$

$$
theta ne left{ {begin{array}{c}
{{0^{text{o}}} + {{360}^{text{o}}}n} \
{{{180}^{text{o}}} + {{360}^{text{o}}}n}
end{array}} right.
$$

where $n$ is an integer

Step 4

4 of 7

b)

L.S.: $tanthetacostheta=dfrac{y}{x}cdot x=y$

R.S.: $sin theta=y$

Restrictions: $cos thetaneq 0$

$$
theta ne left{ {begin{array}{c}
{{90^{text{o}}} + {{360}^{text{o}}}n} \
{{{270}^{text{o}}} + {{360}^{text{o}}}n}
end{array}} right.
$$

where $n$ is an integer

Step 5

5 of 7

c)

L.S.: $csc theta=dfrac{1}{sin theta}=dfrac{1}{y}$

R.S.: $dfrac{1}{sin theta}=dfrac{1}{y}$

Restrictions: $sin thetaneq 0$

$$
theta ne left{ {begin{array}{c}
{{0^{text{o}}} + {{360}^{text{o}}}n} \
{{{180}^{text{o}}} + {{360}^{text{o}}}n}
end{array}} right.
$$

where $n$ is an integer

Step 6

6 of 7

d)

L.S.: $costhetacdot sec theta=xcdot dfrac{1}{x}=1$

R.S.: $1$

Restrictions: $cos thetaneq 0$

$$
theta ne left{ {begin{array}{c}
{{90^{text{o}}} + {{360}^{text{o}}}n} \
{{{270}^{text{o}}} + {{360}^{text{o}}}n}
end{array}} right.
$$

where $n$ is an integer

Result

7 of 7

See proof inside.

Exercise 2

Step 1

1 of 6

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}

Step 2

2 of 6

a) Remember that $(a-b)(a+b)=a^2-b^2$

$(1-sin alpha)(1+sinalpha)=1-sin^2 alpha$

From the Pythagorean identity $sin^2theta+cos^2theta=1$

$$
1-sin^2alpha=cos^2alpha
$$

Step 3

3 of 6

b) Use $tan theta=dfrac{sin theta}{cos theta}$ and $dfrac{1}{cos theta}=sectheta$

$dfrac{tan alpha}{sin alpha}=dfrac{(sin alpha/cosalpha)}{sin alpha}=dfrac{1}{cos alpha}=sec theta$

Step 4

4 of 6

c) Use the Pythogorean identity $sin^2theta+cos^2theta=1$

$$
cos^2alpha+sin^2alpha=1
$$

Step 5

5 of 6

d) Use the quotient identity $cot theta=dfrac{costheta}{sin theta}$

$$
cot alpha cdot sin alpha=dfrac{cosalpha}{sinalpha}cdot sinalpha=cosalpha
$$

Result

6 of 6

a) $cos^2alpha$

b) $sectheta$

c) $1$

d) $cosalpha$

Exercise 3

Step 1

1 of 6

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}

Step 2

2 of 6

a) Remember that $a^2-b^2=(a+b)(a-b)$

$1-cos^2theta$

$=(1)^2-(costheta)^2$

$$
=(1-costheta)(1+costheta)
$$

Step 3

3 of 6

b) Remember that $a^2-b^2=(a+b)(a-b)$

$sin^2theta-cos^2theta$

$=(sin theta)^2-(costheta)^2$

$$
=(sintheta+costheta)(sintheta-costheta)
$$

Step 4

4 of 6

c) Remember that $a^2-2ab+b^2=(a-b)^2$

$sin^2theta-2sintheta+1$

$=(sintheta)^2-2(sintheta)(1)+(1)^2$

$$
=(sintheta-1)^2
$$

Step 5

5 of 6

d) Remember that $a-ab=a(a-b)$

$costheta-cos^2theta$

$$
=costheta(1-costheta)
$$

Result

6 of 6

a) $(1-costheta)(1+costheta)$

b) $(sintheta+costheta)(sintheta-costheta)$

c) $(sintheta-1)^2$

d) $costheta(1-costheta)$

Exercise 4

Step 1

1 of 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}

Step 2

2 of 3

Use $sin^2theta+cos^2theta=1$

L.S: $dfrac{cos^2phi}{1-sinphi}=dfrac{1-sin^2phi}{1-sinphi}$

Factor using $a^2-b^2=(a+b)(a-b)$

$implies dfrac{1-sin^2phi}{1-sinphi}$=$dfrac{(1-sinphi)(1+sinphi)}{1-sinphi}$

$=1+sinphi=$ R.S.

Result

3 of 3

See proof inside.

Exercise 5

Step 1

1 of 7

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}

Step 2

2 of 7

Here, we need to express each ratio in terms of $x$, $y$ and $r$

Remember that for a unit circle, $r=1$

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

$sintheta=dfrac{y}{r}=y$

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

The restrictions on $theta$ are its values that could make any denominator equal to zero.

Step 3

3 of 7

a)
L.S. = $dfrac{sin x}{tan x}=dfrac{sin x}{sin x / cos x}=cos x$ = R.S.

Restrictions: $cos x neq 0$ and $sin x neq 0$

$$
x neq left{ {begin{array}{c}
{{90^{text{o}}} + {{360}^{text{o}}}n} \
{{{270}^{text{o}}} + {{360}^{text{o}}}n} \
{{0^{text{o}}} + {{360}^{text{o}}}n} \
{{{180}^{text{o}}} + {{360}^{text{o}}}n}
end{array}} right.
$$

where $n$ is an integer

Step 4

4 of 7

b)

L.S. = $dfrac{tan theta}{cos theta}=dfrac{sin theta/costheta}{cos theta}=dfrac{sin theta}{cos^2theta}$

Use $sin^2theta+cos^2theta=1$

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

Restrictions: $cos thetaneq 0$

$$
theta ne left{ {begin{array}{c}
{{90^{text{o}}} + {{360}^{text{o}}}n} \
{{{270}^{text{o}}} + {{360}^{text{o}}}n}
end{array}} right.
$$

where $n$ is an integer

Step 5

5 of 7

c)

R.S. = $dfrac{1+sin alpha}{cosalpha}=dfrac{1}{cosalpha}+dfrac{sinalpha}{cosalpha}$

Use $tan theta=dfrac{sintheta}{costheta}$

$dfrac{1}{cosalpha}+dfrac{sinalpha}{cosalpha}=dfrac{1}{cosalpha}+tan alpha =$ L.S.

Restrictions: $cosalpha neq 0$

$$
alpha ne left{ {begin{array}{c}
{{90^{text{o}}} + {{360}^{text{o}}}n} \
{{{270}^{text{o}}} + {{360}^{text{o}}}n}
end{array}} right.
$$

where $n$ is an integer

Step 6

6 of 7

d)
Use $tan theta=dfrac{sintheta}{costheta}$

R.S. = $sintheta costheta tan theta=sintheta costheta cdot dfrac{sin theta}{costheta}= sin^2 theta$

Use $sin^2theta+cos^2theta=1$

$sin^2theta=1-cos^2theta =$ L.S.

Restrictions: $cos xneq 0$

$$
x neq left{ {begin{array}{c}
{{90^{text{o}}} + {{360}^{text{o}}}n} \
{{{270}^{text{o}}} + {{360}^{text{o}}}n}
end{array}} right.
$$

where $n$ is an integer

Result

7 of 7

See proof inside.

Exercise 6

Step 1

1 of 2

Marcia’s reasoning is wrong. An identity must be true for any value of $theta$. You have to show that the both sides are equal in any angle using known fundamental identities. Giving examples of $theta$ that makes the equation is true does NOT prove an identity.

Result

2 of 2

Marcia’s reasoning is incorrect.

Exercise 7

Step 1

1 of 6

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}

Step 2

2 of 6

a)

$sin theta cot theta – sin theta cos theta$

$color{#c34632}{ text{Use};; cottheta=dfrac{costheta}{sintheta}}$

$=sin theta cdot dfrac{costheta}{sintheta}-sinthetacostheta$

$=costheta-sinthetacostheta$

$$
=costheta(1-sintheta)
$$

Step 3

3 of 6

b)

$costheta(1+sectheta)(costheta-1)$

$=costhetaleft(1+dfrac{1}{costheta}right)(costheta-1)$

$=(costheta+1)(costheta-1)$

$color{#c34632}{text{Use} ;; (a+b)(a-b)=a^2-b^2}$

$=cos^2theta-1$

$color{#c34632}{text{Use};; sin^2theta+cos^2theta=1}$

$$
=-sin^2theta
$$

Step 4

4 of 6

c)

$color{#c34632}{text{Use};; (a+b)(a-b)=a^2-b^2}$

$(sin x+cos x)(sin x-cos x)+2cos^2 x$

$=sin^2 x -cos^2 x+2cos^2 x$

$=sin^2 x+cos^2x$

$color{#c34632}{text{Use};; sin^2theta+cos^2theta=1}$

$$
=1
$$

Step 5

5 of 6

d)

$color{#c34632}{text{Let } x=csctheta}$

$dfrac{csc^2theta-3csctheta+2}{csc^2theta-1}=dfrac{x^2-3x+2}{x^2-1}$

$color{#c34632}{text{Factor the numerator and denominator}}$

$=dfrac{(x-1)(x-2)}{(x-1)(x+1)}=dfrac{x-2}{x+1}$

$color{#c34632}text{Revert back to } x=csctheta$

$$
=dfrac{csctheta-2}{csctheta+1}
$$

Result

6 of 6

a) $costheta(1-sintheta)$

b) $-sin^2theta$

c) 1

d) $dfrac{csctheta-2}{csctheta+1}$

Exercise 8

Step 1

1 of 8

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}

Step 2

2 of 8

a)

$color{#c34632} text{Use } sin^2theta+cos^2theta=1$

L.S. = $dfrac{sin^2phi}{1-cosphi}=dfrac{1-cos^2phi}{1-cosphi}$

$color{#c34632}text{Factor using } a^2-b^2=(a+b)(a-b)$

$=dfrac{(1+cosphi)(1-cos phi)}{1-cos phi}=1+cos phi$ = R.S.

Step 3

3 of 8

b)

$color{#c34632}text{Use } tan^2theta+1=sec^2theta$

L.S. = $dfrac{tan^2alpha}{1+tan^2alpha}=dfrac{tan^2alpha}{sec^2alpha}=left(dfrac{tan alpha}{sec alpha}right)^2$

$color{#c34632}text{Use } tan theta=dfrac{sin theta}{costheta} ;text{and};sectheta=dfrac{1}{costheta}$

$=left(dfrac{sin alpha / cos alpha}{1/cosalpha}right)^2=sin^2alpha$ = R.S.

Step 4

4 of 8

c)

$color{#c34632};text{Use } (a-b)(a+b)=a^2-b^2;text{and}; sin^2theta+cos^2theta=1$

R.S. = $(1-sin x)(1+sin x)=1-sin^2 x =cos^2 x$ = L.S.

Step 5

5 of 8

d) $color{#c34632}text{Use } sin^2theta+cos^2theta=1$

$sin^2theta+2cos^2theta-1$

$=(1-cos^2theta)+2cos^2theta-1$

$=1+cos^2theta-1$

$=cos^2theta$ = R.S.

Step 6

6 of 8

e) $color{#c34632};text{Use}; a^4-b^4=(a^2+b^2)(a^2-b^2)$

L.S. = $sin^4alpha-cos^4alpha$

$=(sin^2alpha+cos^2alpha)(sin^2alpha-cos^2alpha)$

$color{#c34632};text{Use}; sin^2theta+cos^2theta=1$

$=(1)(sin^2alpha-cos^2alpha)=$ R.S.

Step 7

7 of 8

f) L.S. = $tan theta+dfrac{1}{tan theta}$

$color{#c34632}text{Use the quotient identities for tangent}$

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

$color{#c34632}text{Use} sin^2theta+cos^2theta=1$

$=dfrac{sin^2theta+cos^2theta}{sintheta costheta}$

$=dfrac{1}{sinthetacostheta}$ = R.S.

Result

8 of 8

See proof inside.

Exercise 9

Step 1

1 of 2

a) It could work but not for all cases such as when there are trigonometric ratios in the denominator. In this case, some parts of the graph would be undefined.

b) This approach can only be used if there are no trigonometric ratios in the denominator.

Result

2 of 2

a) Not for all cases.

b) This approach can only be used if there are no trigonometric ratios in the denominator.

Exercise 10

Step 1

1 of 2

If we can find any value of $theta$ such that both sides are defined but does not make the equation true, then it is NOT an identity.

For $theta=45^circ$, both sides are defined.

L.S. $csc^2 45^circ+sec^2 45^circ =(sqrt{2})^2+(sqrt{2})^2=4$

R.S. $1$

L.S. $neq$ R.S.

Therefore, it is NOT an identity.

Result

2 of 2

not an identity

Exercise 11

Step 1

1 of 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}

Step 2

2 of 3

L.S. = $sin^2 x left(1+dfrac{1}{tan^2 x}right)$

$=sin^2 xleft(dfrac{tan^2x+1}{tan^2x}right)$

$color{#c34632}text{Use }1+tan^2theta=sec^2theta;text{and};sectheta=dfrac{1}{costheta}$

$=sin^2 x left(dfrac{sec^2 x}{tan^2x}right)$

$=sin^2 x left(dfrac{sec x}{tan x}right)^2$

$=sin^2x left(dfrac{1/cos x}{sin x/cos x}right)^2$

$=sin^2xcdot dfrac{1}{sin^2x}$

$=1=$ R.S.

Result

3 of 3

See proof inside.

Exercise 12

Step 1

1 of 4

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}

Step 2

2 of 4

a)

$color{#c34632};text{Use};sin^2theta+cos^2theta=1$

L.S. = $dfrac{sin^2theta+2costheta-1}{sin^2theta+3costheta-3}=dfrac{(1-cos^2theta)+2costheta-1}{(1-cos^2theta)+3costheta-3}$

$=dfrac{-cos^2theta+2costheta}{-cos^2theta+3costheta-2}$

$color{#c34632};text{Let};x=costheta$

$=dfrac{-x^2+2x}{-x^2+3x-2}=dfrac{(-1)(x^2-2x)}{(-1)(x^2-3x+2)}$

$=dfrac{x(x-2)}{(x-1)(x-2)}=dfrac{x}{x-1}=dfrac{costheta}{costheta-1}$

R.S. = $dfrac{cos^2theta+costheta}{-sin^2theta}$

$color{#c34632}text{Use }sin^2theta+cos^2theta=1$

$=dfrac{cos^2theta+costheta}{-(1-cos^2theta)}$

$=dfrac{cos^2theta+costheta}{cos^2theta-1}$

$=dfrac{costheta(costheta+1)}{(costheta+1)(costheta-1)}$

$=dfrac{costheta}{costheta-1}$

$costheta neq 1$ and $sin thetaneq 0$

Step 3

3 of 4

b) R.S. $=dfrac{2sin^2alpha-2sin^4alpha-1}{1-sin^2alpha}$

$color{#c34632}text{Let};y=sin alpha$

$=dfrac{2y^2-2y^4-1}{1-y^2}$

$=dfrac{2y^2(1-y^2)-1}{1-y^2}$

$=dfrac{2y^2(1-y^2)}{1-y^2}-dfrac{1}{1-y^2}$

$=2y^2-dfrac{1}{1-y^2}$

$=2sin^2theta-dfrac{1}{1-sin^2theta}$

$color{#c34632}text{Use};sin^2theta+cos^2theta=1$

$=2sin^2theta-dfrac{1}{cos^2theta}$

$color{#c34632}text{Use};sectheta=dfrac{1}{costheta}$

$=2sin^2theta-sec^2theta$

$color{#c34632}text{Use};tan^2theta+1=sec^2theta$

$=2sin^2theta-(tan^2theta+1)$

$color{#c34632}text{Use};sin^2theta+cos^2theta=1$

$=2sin^2theta-tan^2theta+(sin^2theta+cos^2theta)$

$=sin^2theta+cos^2theta-tan^2theta$ = L.S.

Result

4 of 4

See proof inside.

Exercise 13

Step 1

1 of 5

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}

Step 2

2 of 5

Answers can vary. Here are some examples.

i) Multiply both sides by $dfrac{1}{sintheta}$

$left(dfrac{1}{sintheta}right) (sin^2theta+cos^2theta)=dfrac{1}{sintheta}$

$$
sintheta+csctheta cos^2theta=csctheta
$$

Step 3

3 of 5

ii) Divide both sides by $cos^2theta$

$left(dfrac{1}{cos^2theta}right) (sin^2theta+cos^2theta)=dfrac{1}{cos^2theta}$

$tan^2theta+1=dfrac{1}{cos^2theta}$

Subtract both sides by $1$

$tan^2theta=dfrac{1}{cos^2theta}-1$

Step 4

4 of 5

iii)Multiply both sides by $sectheta$

$left(secthetaright) (sin^2theta+cos^2theta)=1cdot sec theta$

$sin^2thetasectheta+costheta=sectheta$

Result

5 of 5

Answers can vary. See examples inside.

Exercise 14

Step 1

1 of 9

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}

Step 2

2 of 9

[begin{gathered}
{text{i)}} hfill \
hfill \
{text{L}}{text{.S}}{text{.}} = left( {1 – {{cos }^2}x} right)left( {1 – {{tan }^2}x} right){text{ }} hfill \
{text{ }} hfill \
{color{red}text{[Use }}color{red}{color{red}sin ^2}x + {cos ^2}x = 1 Rightarrow 1 – {cos ^2}x = {sin ^2}x] hfill \
hfill \
= {sin ^2}xleft( {1 – {{tan }^2}x} right){text{ }} hfill \
hfill \
color{red} left[color{red} {{text{Use }}color{red}tan x = frac{{sin x }}{{cos x }}} right] hfill \
{text{ }} hfill \
= {sin ^2}xleft( {1 – frac{{{{sin }^2}x}}{{{{cos }^2}x}}} right){text{ }} hfill \
{text{ }} hfill \
= {sin ^2}xleft( {frac{{{{cos }^2}x – {{sin }^2}x}}{{{{cos }^2}x}}} right) hfill \
hfill \
= frac{{{{sin }^2}x{{cos }^2}x – {{sin }^4}x}}{{{{cos }^2}x}} hfill \
hfill \
= frac{{{{sin }^2}x{{cos }^2}x – {{sin }^4}x}}{{{{cos }^2}x}} hfill \
hfill \
{text{color{red}[Use }}color{red}{sin ^2}x + {cos ^2}x = 1 Rightarrow {cos ^2}x = 1 – {sin ^2}x] hfill \
hfill \
= frac{{{{sin }^2}xleft( {1 – {{sin }^2}x} right) – {{sin }^4}x}}{{1 – {{sin }^2}x}} hfill \
{text{ }} hfill \
= frac{{{{sin }^2}x – {{sin }^4}x – {{sin }^4}x}}{{1 – {{sin }^2}x}} hfill \
hfill \
= frac{{{{sin }^2}x – 2{{sin }^4}x}}{{1 – {{sin }^2}x}} = {text{R}}{text{.S}}{text{.}} hfill \
hfill \
end{gathered} ]

Step 3

3 of 9

[begin{gathered}
{text{ii)}} hfill \
1 – 2{cos ^2}phi = {sin ^4}phi – {cos ^4}phi hfill \
hfill \
{text{R}}{text{.S}}{text{. }} = {sin ^4}phi – {cos ^4}phi hfill \
hfill \
color{red} left[ {{text{Factor using }}{a^4} – {b^4} = left( {{a^2} + {b^2}} right)left( {{a^2} – {b^2}} right)} right] hfill \
hfill \
= left( {{{sin }^2}phi + {{cos }^2}phi } right)left( {{{sin }^2}phi – {{cos }^2}phi } right) hfill \
hfill \
color{red} left[ {{text{Use }}{{sin }^2}theta + {{cos }^2}theta = 1} right] hfill \
hfill \
= left( 1 right)left( {{{sin }^2}theta – {{cos }^2}theta } right) hfill \
hfill \
color{red} left[ {{text{Use }}{{sin }^2}theta + {{cos }^2}theta = 1 Rightarrow {{sin }^2}theta = 1 – {{cos }^2}theta } right] hfill \
hfill \
= left( {1 – {{cos }^2}phi } right) – {cos ^2}phi hfill \
hfill \
= 1 – 2{cos ^2}phi = {text{ L}}{text{.S}}{text{.}} hfill \
end{gathered} ]

Step 4

4 of 9

[begin{gathered}
{text{iii)}} hfill \
hfill \
frac{{sin theta tan theta }}{{sin theta + tan theta }} = sin theta tan theta hfill \
hfill \
{text{if }}theta = {text{4}}{{text{5}}^{text{o}}} hfill \
sin {45^{text{o}}} = frac{{sqrt 2 }}{2} hfill \
tan theta = 1 hfill \
hfill \
{text{L}}{text{.S}}{text{.}} = frac{{left( {frac{{sqrt 2 }}{2}} right)left( 1 right)}}{{frac{{sqrt 2 }}{2} + 1}} = frac{{sqrt 2 /2}}{{left( {sqrt 2 + 2} right)/2}} = frac{{sqrt 2 }}{{2 + sqrt 2 }} = 0.414 hfill \
{text{R}}{text{.S}}{text{.}} = left( {frac{{sqrt 2 }}{2}} right)left( 1 right) = frac{{sqrt 2 }}{2} = 0.707 hfill \
hfill \
{text{Thus, it is NOT an identity}}{text{.}} hfill \
end{gathered} ]

Step 5

5 of 9

[begin{gathered}
{text{iv)}} hfill \
hfill \
{text{L}}{text{.S}}{text{.}} = frac{{1 + 2sin beta cos beta }}{{sin beta + cos beta }} hfill \
hfill \
color{red}left[ {{text{Use }}1 = {{sin }^2}beta + {{cos }^2}beta } right] hfill \
hfill \
= frac{{{{sin }^2}beta + {{cos }^2}beta + 2sin beta cos beta }}{{sin beta + cos beta }} hfill \
hfill \
color{red}left[ {{text{Factor using }}left( {{a^2} + 2ab + {b^2}} right) = {{left( {a + b} right)}^2}} right] hfill \
hfill \
= frac{{{{left( {sin beta + cos beta } right)}^2}}}{{sin beta + cos beta }} hfill \
hfill \
= sin beta + cos beta = {text{ R}}{text{.S}}{text{.}} hfill \
end{gathered} ]

Step 6

6 of 9

[begin{gathered}
{text{v)}} hfill \
hfill \
{text{L}}{text{.S}}{text{.}} = frac{{1 – cos beta }}{{sin beta }} hfill \
hfill \
color{red}left[ {{text{Multiply by }}frac{{1 + cos beta }}{{1 + cos beta }}{text{ }}} right] hfill \
hfill \
= frac{{1 – cos beta }}{{sin beta }} cdot frac{{1 + cos beta }}{{1 + cos beta }} hfill \
hfill \
= frac{{1 – {{cos }^2}beta }}{{sin beta left( {1 + cos beta } right)}} hfill \
hfill \
color{red}left[ {{text{Use }}{{sin }^2}beta = 1 – {{cos }^2}beta {text{ }}} right] hfill \
hfill \
= frac{{{{sin }^2}beta }}{{sin beta left( {1 + cos beta } right)}} hfill \
hfill \
= frac{{sin beta }}{{1 + cos beta }} = {text{R}}{text{.S}}{text{.}} hfill \
end{gathered} ]

Step 7

7 of 9

[begin{gathered}
{text{vi)}} hfill \
hfill \
color{red}left[ {{text{Multiply by }}frac{{1 – cos x}}{{1 – cos x}}} right] hfill \
hfill \
{text{L}}{text{.S}}{text{.}} = frac{{sin x}}{{1 + cos x}} cdot frac{{1 – cos x}}{{1 – cos x}} hfill \
hfill \
= frac{{sin xleft( {1 – cos x} right)}}{{1 – {{cos }^2}x}} hfill \
hfill \
color{red} left[ {{text{Use }}1 – {{cos }^2}x = {{sin }^2}x} right] hfill \
hfill \
= frac{{sin xleft( {1 – cos x} right)}}{{{{sin }^2}x}} hfill \
hfill \
= frac{{1 – cos x}}{{sin x}} hfill \
hfill \
color{red}left[ {{text{Use }}frac{1}{{sin x}} = csc x{text{ and }}frac{{cos x}}{{sin x}} = cot x} right] hfill \
hfill \
= csc x – cot x = {text{ R}}{text{.S}}{text{.}} hfill \
end{gathered} ]

Step 8

8 of 9

a) Only option (iii) is not an identity.

b) The restrictions are those that could make the denominator zero:

i) $1-sin^2xneq 0implies$ $sin^2xneq 1$

ii) no restrictions, i.e. true for any real number

iii) not applicable

iv) $sin theta+cos thetaneq 0implies sinthetaneq -costheta$

v) $sin betaneq 0$ and $1+costhetaneq 0implies costhetaneq -1$

vi) $1+cos xneq 0implies cos xneq -1$

Result

9 of 9

See proof inside.

All Solutions

Section 5-5: Trigonometric Identities

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards