Simplify Trigonometric Identities

Topics

Try VividMath Premium to unlock full access

Start Free Trial

Lets see your results!

Points

Score

Time

Retry Quiz

Answered
Review

Question 1 of 7

Which of the following are trigonometric identities?

1. ${sec}^{2} θ = 1 + {sin}^{2} θ$ $\text(sec)^2 theta=1+\text(sin)^2 theta$
2. ${sec}^{2} θ = 1 + {tan}^{2} θ$ $\text(sec)^2 theta=1+\text(tan)^2 theta$
3. ${cot}^{2} θ = {cos}^{2} θ + 1$ $\text(cot)^2 theta=\text(cos)^2 theta+1$
4. ${csc}^{2} θ = {cot}^{2} θ + 1$ $\text(csc)^2 theta=\text(cot)^2 theta +1$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Trigonometric Identities

sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Using the given image, we can see that the radius of the circle will be

1

$1$ . Therefore, the circle follows the formula

x^{2} + y^{2} = 1

$x^2+y^2=1$ , where

1

$1$ is the radius.

Make a reference triangle by making a line from the point of origin

(0, 0)

$(0,0)$ going to any point in the circle to represent the radius and connect it to either the x or y-axis

Notice that, given angle

θ

$theta$ , the circle has the vertical line parallel to the y-axis as its opposite side and the line on the x-axis as the adjacent side

Now that we have the opposite and adjacent sides and the radius

1

$1$ as the hypotenuse, we can use the trigonometric functions to find their respective values

$sin$ $\text(sin)$	$=$ $=$	$\frac{opposite}{hypotenuse}$ $\text(opposite)/(\text(hypotenuse))$

	$=$ $=$	$\frac{y}{1}$ $y/1$

	$=$ $=$	$y$ $y$

$cos$ $\text(cos)$	$=$ $=$	$\frac{adjacent}{hypotenuse}$ $\text(adjacent)/(\text(hypotenuse))$

	$=$ $=$	$\frac{x}{1}$ $x/1$

	$=$ $=$	$x$ $x$

x

$x$ and

y

$y$ are both coordinate values, so we can write the coordinates as

(cos θ, sin θ)

$(\text(cos) theta,\text(sin) theta)$

Finally, use the values of

x

$x$ and

y

$y$ and derive its trigonometric identities

$x^{2} + y^{2}$ $x^2+y^2$	$=$ $=$	$1$ $1$
${cos}^{2} θ + {sin}^{2} θ$ $\text(cos)^2theta+\text(sin)^2theta$	$=$ $=$	$1$ $1$	Substitute known values
$({cos}^{2} θ + {sin}^{2} θ)$ $(\text(cos)^2theta+\text(sin)^2theta)$ $\div {cos}^{2} θ$ $divide \text(cos)^2theta$	$=$ $=$	$1$ $1$ $\div {cos}^{2} θ$ $divide \text(cos)^2theta$	Divide both sides by ${cos}^{2} θ$ $\text(cos)^2theta$

$1 + \frac{{sin}^{2} θ}{{cos}^{2} θ}$ $1+(\text(sin)^2theta)/(\text(cos)^2theta)$	$=$ $=$	$\frac{1}{{cos}^{2} θ}$ $1/(\text(cos)^2theta)$

$1 +$ $1+$ ${tan}^{2} θ$ $\text(tan)^2theta$	$=$ $=$	$\frac{1}{{cos}^{2} θ}$ $1/(\text(cos)^2theta)$	$\frac{{sin}^{2} θ}{{cos}^{2} θ} = {tan}^{2} θ$ $(\text(sin)^2theta)/(\text(cos)^2theta)=\text(tan)^2theta$

$1 + {tan}^{2} θ$ $1+\text(tan)^2theta$	$=$ $=$	${sec}^{2} θ$ $\text(sec)^2theta$	$\frac{1}{{cos}^{2} θ} = {sec}^{2} θ$ $1/(\text(cos)^2theta)=\text(sec)^2theta$
${sec}^{2} θ$ $\text(sec)^2theta$	$=$ $=$	$1 + {tan}^{2} θ$ $1+\text(tan)^2theta$

$x^{2} + y^{2}$ $x^2+y^2$	$=$ $=$	$1$ $1$
${cos}^{2} θ + {sin}^{2} θ$ $\text(cos)^2theta+\text(sin)^2theta$	$=$ $=$	$1$ $1$	Substitute known values
$({cos}^{2} θ + {sin}^{2} θ)$ $(\text(cos)^2theta+\text(sin)^2theta)$ $\div {sin}^{2} θ$ $divide \text(sin)^2theta$	$=$ $=$	$1$ $1$ $\div {sin}^{2} θ$ $divide \text(sin)^2theta$	Divide both sides by ${sin}^{2} θ$ $\text(sin)^2theta$

$\frac{{cos}^{2} θ}{{sin}^{2} θ} + 1$ $(\text(cos)^2theta)/(\text(sin)^2theta)+1$	$=$ $=$	$\frac{1}{{sin}^{2} θ}$ $1/(\text(sin)^2theta)$

${cot}^{2} θ$ $\text(cot)^2theta$ $+ 1$ $+1$	$=$ $=$	$\frac{1}{{sin}^{2} θ}$ $1/(\text(sin)^2theta)$	$\frac{{cos}^{2} θ}{{sin}^{2} θ} = {cot}^{2} θ$ $(\text(cos)^2theta)/(\text(sin)^2theta)=\text(cot)^2theta$

${cot}^{2} θ + 1$ $\text(cot)^2theta+1$	$=$ $=$	${csc}^{2} θ$ $\text(csc)^2theta$	$\frac{1}{{sin}^{2} θ} = {csc}^{2} θ$ $1/(\text(sin)^2theta)=\text(csc)^2theta$
${csc}^{2} θ$ $\text(csc)^2theta$	$=$ $=$	${cot}^{2} θ + 1$ $\text(cot)^2theta+1$

{sec}^{2} θ = 1 + {tan}^{2} θ

$\text(sec)^2theta=1+\text(tan)^2theta$

{csc}^{2} θ = {cot}^{2} θ + 1

$\text(csc)^2theta=\text(cot)^2theta+1$

Question 2 of 7

Simplify

\frac{1}{1 + sin θ} + \frac{1}{1 - sin θ}

$1/(1+sin theta)+1/(1-sin theta)$

1. ${csc}^{2} θ$ $\text(csc)^2theta$
2. ${csc}^{2} θ + 2$ $\text(csc)^2theta+2$
3. $2 {sec}^{2} θ$ $2\text(sec)^2theta$
4. $\frac{2}{{sec}^{2} θ}$ $2/(\text(sec)^2theta)$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Trigonometric Identities

sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Apply the rule of adding fractions and simplify the expression

	$\frac{1}{1 + sin θ} + \frac{1}{1 - sin θ}$ $1/(1+\text(sin) theta)+1/(1-\text(sin) theta)$

$=$ $=$	$\frac{1 - sin θ + 1 + sin θ}{(1 + sin θ) (1 - sin θ)}$ $(1-\text(sin) theta+1+\text(sin) theta)/((1+\text(sin) theta)(1-\text(sin) theta))$

$=$ $=$	$\frac{2}{(1 + sin θ) (1 - sin θ)}$ $2/((1+\text(sin) theta)(1-\text(sin) theta))$	Combine like terms

$=$ $=$	$\frac{2}{1 - {sin}^{2} θ}$ $2/(1-\text(sin)^2theta)$	Expand

Recall that

{sin}^{2} θ + {cos}^{2} θ = 1

$\text(sin)^2theta+\text(cos)^2theta=1$ . Derive this formula to further simplify the expression

${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$	$=$ $=$	$1$ $1$
${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	$=$ $=$	$1$ $1$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	Subtract ${sin}^{2} θ$ $\text(sin)^2theta$ from both sides
${cos}^{2} θ$ $\text(cos)^2theta$	$=$ $=$	$1 - {sin}^{2} θ$ $1-\text(sin)^2theta$

	$\frac{2}{1 - {sin}^{2} θ}$ $2/(1-\text(sin)^2theta)$

$=$ $=$	$\frac{2}{{cos}^{2} θ}$ $2/(\text(cos)^2theta)$	$1 - {sin}^{2} θ = {cos}^{2} θ$ $1-\text(sin)^2theta=\text(cos)^2theta$

$=$ $=$	$\frac{2}{1} \times \frac{1}{{cos}^{2} θ}$ $2/1xx 1/(\text(cos)^2theta)$	Take out $2$ $2$

$=$ $=$	$2 \times {sec}^{2} θ$ $2xx\text(sec)^2theta$	$\frac{1}{{cos}^{2} θ} = {sec}^{2} θ$ $1/(\text(cos)^2theta)=\text(sec)^2theta$
$=$ $=$	$2 {sec}^{2} θ$ $2\text(sec)^2theta$

2 {sec}^{2} θ

$2\text(sec)^2theta$

Question 3 of 7

Simplify

\frac{tan θ}{sec θ}

$(tan theta)/(sec theta)$

1. $sin θ$ $\text(sin) theta$
2. $cos θ$ $\text(cos) theta$
3. $\frac{1}{tan θ}$ $1/(\text(tan) theta)$
4. $2 sec θ$ $2\text(sec) theta$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Trigonometric Identities

sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Substitute known values to

tan

$\text(tan)$ and

sec

$\text(sec)$

$tan$ $\text(tan)$	$=$ $=$	$\frac{sin}{cos}$ $(\text(sin))/(\text(cos))$

$sec$ $\text(sec)$	$=$ $=$	$\frac{1}{cos}$ $1/(\text(cos))$

	$\frac{tan θ}{sec θ}$ $(\text(tan) theta)/(\text(sec) theta)$

$=$ $=$	$\frac{\frac{sin}{cos}}{\frac{1}{cos}}$ $((\text(sin))/(\text(cos)))/(1/(\text(cos)))$	Substitute known values

$=$ $=$	$sin θ$ $\text(sin) theta$	Apply rule of dividing fractions

sin θ

$\text(sin) theta$

Question 4 of 7

Simplify

sec θ - sec θ {sin}^{2} θ

$sec theta-sec theta sin^2theta$

1. $cos θ$ $\text(cos) theta$
2. $sin θ$ $\text(sin) theta$
3. $csc θ$ $\text(csc) theta$
4. $sec θ$ $\text(sec) theta$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Trigonometric Identities

sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Factor out

sec θ

$\text(sec) theta$ on both terms

		$sec θ - sec θ {sin}^{2} θ$ $\text(sec) theta-\text(sec) theta \text(sin)^2theta$
	$=$ $=$	$sec θ (1 - {sin}^{2} θ)$ $\text(sec) theta(1-\text(sin)^2theta)$	Factorise

Recall that

{sin}^{2} θ + {cos}^{2} θ = 1

$\text(sin)^2theta+\text(cos)^2theta=1$ . Derive this formula to further simplify the expression

${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$	$=$ $=$	$1$ $1$
${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	$=$ $=$	$1$ $1$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	Subtract ${sin}^{2} θ$ $\text(sin)^2theta$ from both sides
${cos}^{2} θ$ $\text(cos)^2theta$	$=$ $=$	$1 - {sin}^{2} θ$ $1-\text(sin)^2theta$

	$sec θ (1 - {sin}^{2} θ)$ $\text(sec) theta(1-\text(sin)^2theta)$
$=$ $=$	$sec θ ({cos}^{2} θ)$ $\text(sec) theta(\text(cos)^2theta)$	$1 - {sin}^{2} θ = {cos}^{2} θ$ $1-\text(sin)^2theta=\text(cos)^2theta$

$=$ $=$	$\frac{1}{cos θ} \cdot ({cos}^{2} θ)$ $1/(\text(cos) theta)*(\text(cos)^2theta)$	$sec θ = \frac{1}{cos θ}$ $\text(sec) theta=1/(\text(cos) theta)$

$=$ $=$	$\frac{{cos}^{2} θ}{cos θ}$ $(\text(cos)^2theta)/(\text(cos) theta)$

$=$ $=$	$cos θ$ $\text(cos) theta$

cos θ

$\text(cos) theta$

Question 5 of 7

5. Question
Simplify

${sec}^{2} A - {csc}^{2} A$ $sec^2A-csc^2A$
- 1. $(sec A - csc A) (sec A + csc A)$ $(\text(sec) A-\text(csc) A)(\text(sec) A+\text(csc) A)$
- 2. $(sec A - csc A) (sec A - csc A)$ $(\text(sec) A-\text(csc) A)(\text(sec) A-\text(csc) A)$
- 3. $(sec A + csc A) (sec A + csc A)$ $(\text(sec) A+\text(csc) A)(\text(sec) A+\text(csc) A)$
- 4. $A (sec A + csc A)$ $A(\text(sec) A+\text(csc) A)$
Hint

Help Video
Correct

Keep Going!
Incorrect
Need Text
Play
Current Time 0:00
/
Duration Time 0:00
Remaining Time -0:00
Stream TypeLIVE
Loaded: 0%
Progress: 0%
0:00
Fullscreen
Video Acceleration:
On
Off

00:00
Mute
Playback Rate
1x
2x
1.5x
1.25x
1x
0.75x
0.5x
Subtitles
subtitles off
Captions
captions off
English
Chapters
Chapters

The difference of two squares is a format of a quadratic expression where there are two terms that are both perfect squares.

First, find the square of the two terms.

First term

${sec}^{2} A$ $\text(sec)^2A$ $=$ $=$ $sec A \cdot sec A$ $\text(sec) A*\text(sec) A$

Second term

${csc}^{2} A$ $\text(csc)^2A$ $=$ $=$ $csc A \cdot csc A$ $\text(csc) A*\text(csc) A$

Finally, get the sum and difference of the two values.

$($ $($ $sec A$ $\text(sec) A$ $-$ $-$ $csc A$ $\text(csc) A$ $) ($ $)($ $sec A$ $\text(sec) A$ $+$ $+$ $csc A$ $\text(csc) A$ $)$ $)$

$(sec A - csc A) (sec A + csc A)$ $(\text(sec) A-\text(csc) A)(\text(sec) A+\text(csc) A)$

Question 6 of 7

Simplify

\frac{{tan}^{2} θ}{sec θ - 1}

$(tan^2theta)/(sec theta-1)$

1. $\frac{1}{sec θ}$ $1/(\text(sec)theta)$
2. $sec θ - 1$ $\text(sec)theta-1$
3. $- \frac{1}{sec θ}$ $-1/(\text(sec)theta)$
4. $sec θ + 1$ $\text(sec)theta+1$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Trigonometric Identities

sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Recall that

1 + {tan}^{2} θ = {sec}^{2} θ

$1+\text(tan)^2theta=\text(sec)^2theta$ . Derive this formula to further simplify the expression

$1 + {tan}^{2} θ$ $1+\text(tan)^2theta$	$=$ $=$	${sec}^{2} θ$ $\text(sec)^2theta$
$1 + {tan}^{2} θ$ $1+\text(tan)^2theta$ $- 1$ $-1$	$=$ $=$	${sec}^{2} θ$ $\text(sec)^2theta$ $- 1$ $-1$	Subtract $1$ $1$ from both sides
${tan}^{2} θ$ $\text(tan)^2theta$	$=$ $=$	${sec}^{2} θ - 1$ $\text(sec)^2theta-1$

	$\frac{{tan}^{2} θ}{sec θ - 1}$ $(\text(tan)^2theta)/(\text(sec) theta-1)$

$=$ $=$	$\frac{{sec}^{2} θ - 1}{sec θ - 1}$ $(\text(sec)^2theta-1)/(\text(sec) theta-1)$	${tan}^{2} θ = {sec}^{2} θ - 1$ $\text(tan)^2theta=\text(sec)^2theta-1$

$=$ $=$	$\frac{(sec θ - 1) (sec θ + 1)}{sec θ - 1}$ $((\text(sec)theta-1)(\text(sec)theta+1))/(\text(sec) theta-1)$	Factorise

$=$ $=$	$sec θ + 1$ $\text(sec)theta+1$	$\frac{sec θ - 1}{sec θ - 1} = 1$ $(\text(sec) theta-1)/(\text(sec) theta-1)=1$

sec θ + 1

$\text(sec)theta+1$

Question 7 of 7

Simplify

\frac{({tan}^{2} β) ({cot}^{2} β + 1)}{{tan}^{2} β + 1}

$((tan^2beta)(cot^2beta+1))/(tan^2beta+1)$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Trigonometric Identities

sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Recall that

1 + {cot}^{2} θ = {csc}^{2} θ

$1+\text(cot)^2theta=\text(csc)^2theta$ . Derive this formula to further simplify the expression

	$\frac{({tan}^{2} β) ({cot}^{2} β + 1)}{{tan}^{2} β + 1}$ $((\text(tan)^2beta)(\text(cot)^2beta+1))/(\text(tan)^2beta+1)$

$=$ $=$	$\frac{{tan}^{2} β \cdot {csc}^{2} θ}{{tan}^{2} β + 1}$ $(\text(tan)^2beta*\text(csc)^2theta)/(\text(tan)^2beta+1)$	$1 + {cot}^{2} θ = {csc}^{2} θ$ $1+\text(cot)^2theta=\text(csc)^2theta$

Finally, substitute known values to

tan

$\text(tan)$ ,

csc

$\text(csc)$ , and

sec

$\text(sec)$

$tan$ $\text(tan)$	$=$ $=$	$\frac{sin}{cos}$ $(\text(sin))/(\text(cos))$

$csc$ $\text(csc)$	$=$ $=$	$\frac{1}{sin}$ $1/(\text(sin))$

$sec$ $\text(sec)$	$=$ $=$	$\frac{1}{cos}$ $1/(\text(cos))$

	$\frac{{tan}^{2} β \cdot {csc}^{2} θ}{{tan}^{2} β + 1}$ $(\text(tan)^2beta*\text(csc)^2theta)/(\text(tan)^2beta+1)$

$=$ $=$	$\frac{\frac{{sin}^{2} β}{{cos}^{2} β} \cdot \frac{1}{{csc}^{2} θ}}{\frac{1}{{cos}^{2} θ}}$ $((\text(sin)^2beta)/(\text(cos)^2beta)*1/(\text(csc)^2theta))/(1/(\text(cos)^2theta))$	Substitute known values

$=$ $=$	$\frac{{sin}^{2} β}{{cos}^{2} β} \cdot \frac{1}{{csc}^{2} θ} \cdot \frac{{cos}^{2} θ}{1}$ $(\text(sin)^2beta)/(\text(cos)^2beta)1/(\text(csc)^2theta)(\text(cos)^2theta)/1$	Apply rule of dividing fractions

$=$ $=$	$1$ $1$

1

$1$

Simplify Trigonometric Identities

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Well Done!

Trigonometric Identities

2. Question

Hint

Nice Job!

Trigonometric Identities

3. Question

Hint

Excellent!

Trigonometric Identities

4. Question

Hint

Fantastic!

Trigonometric Identities

5. Question

Hint

Keep Going!

6. Question

Hint

Correct!

Trigonometric Identities

7. Question

Hint

Great Work!

Trigonometric Identities

Quizzes