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Professor Murray

Professor Murray

Identity Tan(squared)x+1=Sec(squared)x

Slide Duration:

Table of Contents

I. Trigonometric Functions
Angles

39m 5s

Intro
0:00
Degrees
0:22
Circle is 360 Degrees
0:48
Splitting a Circle
1:13
Radians
2:08
Circle is 2 Pi Radians
2:31
One Radian
2:52
Half-Circle and Right Angle
4:00
Converting Between Degrees and Radians
6:24
Formulas for Degrees and Radians
6:52
Coterminal, Complementary, Supplementary Angles
7:23
Coterminal Angles
7:30
Complementary Angles
9:40
Supplementary Angles
10:08
Example 1: Dividing a Circle
10:38
Example 2: Converting Between Degrees and Radians
11:56
Example 3: Quadrants and Coterminal Angles
14:18
Extra Example 1: Common Angle Conversions
-1
Extra Example 2: Quadrants and Coterminal Angles
-2
Sine and Cosine Functions

43m 16s

Intro
0:00
Sine and Cosine
0:15
Unit Circle
0:22
Coordinates on Unit Circle
1:03
Right Triangles
1:52
Adjacent, Opposite, Hypotenuse
2:25
Master Right Triangle Formula: SOHCAHTOA
2:48
Odd Functions, Even Functions
4:40
Example: Odd Function
4:56
Example: Even Function
7:30
Example 1: Sine and Cosine
10:27
Example 2: Graphing Sine and Cosine Functions
14:39
Example 3: Right Triangle
21:40
Example 4: Odd, Even, or Neither
26:01
Extra Example 1: Right Triangle
-1
Extra Example 2: Graphing Sine and Cosine Functions
-2
Sine and Cosine Values of Special Angles

33m 5s

Intro
0:00
45-45-90 Triangle and 30-60-90 Triangle
0:08
45-45-90 Triangle
0:21
30-60-90 Triangle
2:06
Mnemonic: All Students Take Calculus (ASTC)
5:21
Using the Unit Circle
5:59
New Angles
6:21
Other Quadrants
9:43
Mnemonic: All Students Take Calculus
10:13
Example 1: Convert, Quadrant, Sine/Cosine
13:11
Example 2: Convert, Quadrant, Sine/Cosine
16:48
Example 3: All Angles and Quadrants
20:21
Extra Example 1: Convert, Quadrant, Sine/Cosine
-1
Extra Example 2: All Angles and Quadrants
-2
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D

52m 3s

Intro
0:00
Amplitude and Period of a Sine Wave
0:38
Sine Wave Graph
0:58
Amplitude: Distance from Middle to Peak
1:18
Peak: Distance from Peak to Peak
2:41
Phase Shift and Vertical Shift
4:13
Phase Shift: Distance Shifted Horizontally
4:16
Vertical Shift: Distance Shifted Vertically
6:48
Example 1: Amplitude/Period/Phase and Vertical Shift
8:04
Example 2: Amplitude/Period/Phase and Vertical Shift
17:39
Example 3: Find Sine Wave Given Attributes
25:23
Extra Example 1: Amplitude/Period/Phase and Vertical Shift
-1
Extra Example 2: Find Cosine Wave Given Attributes
-2
Tangent and Cotangent Functions

36m 4s

Intro
0:00
Tangent and Cotangent Definitions
0:21
Tangent Definition
0:25
Cotangent Definition
0:47
Master Formula: SOHCAHTOA
1:01
Mnemonic
1:16
Tangent and Cotangent Values
2:29
Remember Common Values of Sine and Cosine
2:46
90 Degrees Undefined
4:36
Slope and Menmonic: ASTC
5:47
Uses of Tangent
5:54
Example: Tangent of Angle is Slope
6:09
Sign of Tangent in Quadrants
7:49
Example 1: Graph Tangent and Cotangent Functions
10:42
Example 2: Tangent and Cotangent of Angles
16:09
Example 3: Odd, Even, or Neither
18:56
Extra Example 1: Tangent and Cotangent of Angles
-1
Extra Example 2: Tangent and Cotangent of Angles
-2
Secant and Cosecant Functions

27m 18s

Intro
0:00
Secant and Cosecant Definitions
0:17
Secant Definition
0:18
Cosecant Definition
0:33
Example 1: Graph Secant Function
0:48
Example 2: Values of Secant and Cosecant
6:49
Example 3: Odd, Even, or Neither
12:49
Extra Example 1: Graph of Cosecant Function
-1
Extra Example 2: Values of Secant and Cosecant
-2
Inverse Trigonometric Functions

32m 58s

Intro
0:00
Arcsine Function
0:24
Restrictions between -1 and 1
0:43
Arcsine Notation
1:26
Arccosine Function
3:07
Restrictions between -1 and 1
3:36
Cosine Notation
3:53
Arctangent Function
4:30
Between -Pi/2 and Pi/2
4:44
Tangent Notation
5:02
Example 1: Domain/Range/Graph of Arcsine
5:45
Example 2: Arcsin/Arccos/Arctan Values
10:46
Example 3: Domain/Range/Graph of Arctangent
17:14
Extra Example 1: Domain/Range/Graph of Arccosine
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
Computations of Inverse Trigonometric Functions

31m 8s

Intro
0:00
Inverse Trigonometric Function Domains and Ranges
0:31
Arcsine
0:41
Arccosine
1:14
Arctangent
1:41
Example 1: Arcsines of Common Values
2:44
Example 2: Odd, Even, or Neither
5:57
Example 3: Arccosines of Common Values
12:24
Extra Example 1: Arctangents of Common Values
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
II. Trigonometric Identities
Pythagorean Identity

19m 11s

Intro
0:00
Pythagorean Identity
0:17
Pythagorean Triangle
0:27
Pythagorean Identity
0:45
Example 1: Use Pythagorean Theorem to Prove Pythagorean Identity
1:14
Example 2: Find Angle Given Cosine and Quadrant
4:18
Example 3: Verify Trigonometric Identity
8:00
Extra Example 1: Use Pythagorean Identity to Prove Pythagorean Theorem
-1
Extra Example 2: Find Angle Given Cosine and Quadrant
-2
Identity Tan(squared)x+1=Sec(squared)x

23m 16s

Intro
0:00
Main Formulas
0:19
Companion to Pythagorean Identity
0:27
For Cotangents and Cosecants
0:52
How to Remember
0:58
Example 1: Prove the Identity
1:40
Example 2: Given Tan Find Sec
3:42
Example 3: Prove the Identity
7:45
Extra Example 1: Prove the Identity
-1
Extra Example 2: Given Sec Find Tan
-2
Addition and Subtraction Formulas

52m 52s

Intro
0:00
Addition and Subtraction Formulas
0:09
How to Remember
0:48
Cofunction Identities
1:31
How to Remember Graphically
1:44
Where to Use Cofunction Identities
2:52
Example 1: Derive the Formula for cos(A-B)
3:08
Example 2: Use Addition and Subtraction Formulas
16:03
Example 3: Use Addition and Subtraction Formulas to Prove Identity
25:11
Extra Example 1: Use cos(A-B) and Cofunction Identities
-1
Extra Example 2: Convert to Radians and use Formulas
-2
Double Angle Formulas

29m 5s

Intro
0:00
Main Formula
0:07
How to Remember from Addition Formula
0:18
Two Other Forms
1:35
Example 1: Find Sine and Cosine of Angle using Double Angle
3:16
Example 2: Prove Trigonometric Identity using Double Angle
9:37
Example 3: Use Addition and Subtraction Formulas
12:38
Extra Example 1: Find Sine and Cosine of Angle using Double Angle
-1
Extra Example 2: Prove Trigonometric Identity using Double Angle
-2
Half-Angle Formulas

43m 55s

Intro
0:00
Main Formulas
0:09
Confusing Part
0:34
Example 1: Find Sine and Cosine of Angle using Half-Angle
0:54
Example 2: Prove Trigonometric Identity using Half-Angle
11:51
Example 3: Prove the Half-Angle Formula for Tangents
18:39
Extra Example 1: Find Sine and Cosine of Angle using Half-Angle
-1
Extra Example 2: Prove Trigonometric Identity using Half-Angle
-2
III. Applications of Trigonometry
Trigonometry in Right Angles

25m 43s

Intro
0:00
Master Formula for Right Angles
0:11
SOHCAHTOA
0:15
Only for Right Triangles
1:26
Example 1: Find All Angles in a Triangle
2:19
Example 2: Find Lengths of All Sides of Triangle
7:39
Example 3: Find All Angles in a Triangle
11:00
Extra Example 1: Find All Angles in a Triangle
-1
Extra Example 2: Find Lengths of All Sides of Triangle
-2
Law of Sines

56m 40s

Intro
0:00
Law of Sines Formula
0:18
SOHCAHTOA
0:27
Any Triangle
0:59
Graphical Representation
1:25
Solving Triangle Completely
2:37
When to Use Law of Sines
2:55
ASA, SAA, SSA, AAA
2:59
SAS, SSS for Law of Cosines
7:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
8:44
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:30
Example 3: How Many Triangles Satisfy Conditions, Solve Completely
28:32
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: How Many Triangles Satisfy Conditions, Solve Completely
-2
Law of Cosines

49m 5s

Intro
0:00
Law of Cosines Formula
0:23
Graphical Representation
0:34
Relates Sides to Angles
1:00
Any Triangle
1:20
Generalization of Pythagorean Theorem
1:32
When to Use Law of Cosines
2:26
SAS, SSS
2:30
Heron's Formula
4:49
Semiperimeter S
5:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
5:53
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:19
Example 3: Find Area of a Triangle Given All Side Lengths
26:33
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: Length of Third Side and Area of Triangle
-2
Finding the Area of a Triangle

27m 37s

Intro
0:00
Master Right Triangle Formula and Law of Cosines
0:19
SOHCAHTOA
0:27
Law of Cosines
1:23
Heron's Formula
2:22
Semiperimeter S
2:37
Example 1: Area of Triangle with Two Sides and One Angle
3:12
Example 2: Area of Triangle with Three Sides
6:11
Example 3: Area of Triangle with Three Sides, No Heron's Formula
8:50
Extra Example 1: Area of Triangle with Two Sides and One Angle
-1
Extra Example 2: Area of Triangle with Two Sides and One Angle
-2
Word Problems and Applications of Trigonometry

34m 25s

Intro
0:00
Formulas to Remember
0:11
SOHCAHTOA
0:15
Law of Sines
0:55
Law of Cosines
1:48
Heron's Formula
2:46
Example 1: Telephone Pole Height
4:01
Example 2: Bridge Length
7:48
Example 3: Area of Triangular Field
14:20
Extra Example 1: Kite Height
-1
Extra Example 2: Roads to a Town
-2
Vectors

46m 42s

Intro
0:00
Vector Formulas and Concepts
0:12
Vectors as Arrows
0:28
Magnitude
0:38
Direction
0:50
Drawing Vectors
1:16
Uses of Vectors: Velocity, Force
1:37
Vector Magnitude Formula
3:15
Vector Direction Formula
3:28
Vector Components
6:27
Example 1: Magnitude and Direction of Vector
8:00
Example 2: Force to a Box on a Ramp
12:25
Example 3: Plane with Wind
18:30
Extra Example 1: Components of a Vector
-1
Extra Example 2: Ship with a Current
-2
IV. Complex Numbers and Polar Coordinates
Polar Coordinates

1h 7m 35s

Intro
0:00
Polar Coordinates vs Rectangular/Cartesian Coordinates
0:12
Rectangular Coordinates, Cartesian Coordinates
0:23
Polar Coordinates
0:59
Converting Between Polar and Rectangular Coordinates
2:06
R
2:16
Theta
2:48
Example 1: Convert Rectangular to Polar Coordinates
6:53
Example 2: Convert Polar to Rectangular Coordinates
17:28
Example 3: Graph the Polar Equation
28:00
Extra Example 1: Convert Polar to Rectangular Coordinates
-1
Extra Example 2: Graph the Polar Equation
-2
Complex Numbers

35m 59s

Intro
0:00
Main Definition
0:07
Number i
0:23
Complex Number Form
0:33
Powers of Imaginary Number i
1:00
Repeating Pattern
1:43
Operations on Complex Numbers
3:30
Adding and Subtracting Complex Numbers
3:39
Multiplying Complex Numbers
4:39
FOIL Method
5:06
Conjugation
6:29
Dividing Complex Numbers
7:34
Conjugate of Denominator
7:45
Example 1: Solve For Complex Number z
11:02
Example 2: Expand and Simplify
15:34
Example 3: Simplify the Powers of i
17:50
Extra Example 1: Simplify
-1
Extra Example 2: All Complex Numbers Satisfying Equation
-2
Polar Form of Complex Numbers

40m 43s

Intro
0:00
Polar Coordinates
0:49
Rectangular Form
0:52
Polar Form
1:25
R and Theta
1:51
Polar Form Conversion
2:27
R and Theta
2:35
Optimal Values
4:05
Euler's Formula
4:25
Multiplying Two Complex Numbers in Polar Form
6:10
Multiply r's Together and Add Exponents
6:32
Example 1: Convert Rectangular to Polar Form
7:17
Example 2: Convert Polar to Rectangular Form
13:49
Example 3: Multiply Two Complex Numbers
17:28
Extra Example 1: Convert Between Rectangular and Polar Forms
-1
Extra Example 2: Simplify Expression to Polar Form
-2
DeMoivre's Theorem

57m 37s

Intro
0:00
Introduction to DeMoivre's Theorem
0:10
n nth Roots
3:06
DeMoivre's Theorem: Finding nth Roots
3:52
Relation to Unit Circle
6:29
One nth Root for Each Value of k
7:11
Example 1: Convert to Polar Form and Use DeMoivre's Theorem
8:24
Example 2: Find Complex Eighth Roots
15:27
Example 3: Find Complex Roots
27:49
Extra Example 1: Convert to Polar Form and Use DeMoivre's Theorem
-1
Extra Example 2: Find Complex Fourth Roots
-2
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Lecture Comments (13)

1 answer

Last reply by: Dr. William Murray
Thu Mar 27, 2014 5:00 PM

Post by Tami Cummins on March 22, 2014

in example 3 at 12:46 how is 1/cos + 1 = 1 + cos?  

2 answers

Last reply by: Dr. William Murray
Mon Oct 21, 2013 10:42 PM

Post by Heinz Krug on October 21, 2013

Example III can be used to show the application of the identity tan^2(θ)+1 = sec^2(θ), but it also shows that sometimes there are simpler solutions, like e.g. this one:

LHS = (cscθ-cotθ)/(secθ-1) = (1/sinθ-cosθ/sinθ)/(1/cosθ-1) =
= ((1-cosθ)/sinθ)/((1-cosθ)/cosθ) = (1/sinθ)/(1/cosθ) =
= cosθ/sinθ = cotθ = RHS;

1 answer

Last reply by: Dr. William Murray
Sat Jun 8, 2013 6:01 PM

Post by William Davis on June 6, 2013

You are by far the best lecturer this site has. You actually care and put forth effort unlike a lot of profs on this site. I wish you taught the Cal I section.

1 answer

Last reply by: Dr. William Murray
Sun May 12, 2013 5:40 PM

Post by William Davis on December 28, 2011

In example III he's using strategies I ever knew you could do, which i believe makes solving complex identities treacherous. You should be able to still solve it without looking for shortcuts. but when I try, I am unable to. It's very frustrating.

1 answer

Last reply by: Dr. William Murray
Sun May 12, 2013 5:33 PM

Post by Jonathan Taylor on October 26, 2011

Ex.3 IS JUST CONFUSING IS THERE A SIMPLER FORM

1 answer

Last reply by: Dr. William Murray
Sun May 12, 2013 5:30 PM

Post by Hero Miles on May 16, 2011

I converted example III into sines and cosines first. It was much easier.

Identity Tan(squared)x+1=Sec(squared)x

Main formulas:

  • For any angle x for which the tangent and secant are defined, we have tan2 x + 1 = sec2 x.
  • For any angle x for which the tangent and secant are defined, we have cot2 x + 1 = csc2 x.

Example 1:

Prove the identity tan2 x + 1 = sec2 x.

Example 2:

Given that tanθ = − 4.21 and (π/2) < θ< π , find secθ .

Example 3:

Prove the following trigonometric identity:
cscθ− cotθ

secθ− 1
= cotθ

Example 4:

Prove the identity cot2 x + 1 = csc2 x.

Example 5:

Given that secθ = (13/12) and 270° < θ < 360° , find tanθ .

Identity Tan(squared)x+1=Sec(squared)x

Given that tanθ = - 5.3 and [(3π)/2] <θ < 2π, find secθ.
  • Use the Pythagorean Identity to solve for secθ: 1 + tan2θ = sec2θ
  • 1 + ( − 5.31)2 = sec2θ ⇒ 1 + 28.1961 = sec2θ ⇒ sec2θ = 29.1961 ⇒ secθ = ±√{29.1961}
  • secθ = ± 5.4 Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV which means the same holds true for secant values.
secθ = 5.4 because θ is in quadrant IV
Given that secθ = [13/5] and 180°<θ < 270° , find tanθ.
  • Use the Pythagorean Identity to solve for tanθ: 1 + tan2θ = sec2θ
  • 1 + tan2θ = ( [13/5] )2 ⇒ 1 + tan2θ = [169/25] ⇒ tan2θ = [169/25] - [25/25] ⇒ tanθ = ±√{[144/25]}
  • tanθ = ±[12/5] Remember the mnemonic ASTC. This tells which quadrant tangent values will be positive. Tangent is only positive in quadrants I and III.
tanθ = [12/5] because θ is in quadrant III
Given that tanθ = - 3.23 and [(π)/2] <θ <π, find secθ.
  • Use the Pythagorean Identity to solve for secθ: 1 + tan2θ = sec2θ
  • 1 + ( − 3.23)2 = sec2θ ⇒ 1 + 10.4329 = sec2θ⇒ sec2θ = 11.4329 ⇒ secθ = ±√{11.4329}
  • secθ = ± 3.38126 Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV which means the same holds true for secant values.
secθ = - 3.4 because θ is in quadrant II
Given that secθ = − [5/4] and 90°<θ < 180°, find tanθ.
  • Use the Pythagorean Identity to solve for tanθ: 1 + tan2θ = sec2θ
  • 1 + tan2θ = ( − [5/4] )2 ⇒ 1 + tan2θ = [25/16] ⇒ tan2θ = [25/16] - [16/16] ⇒ tanθ = ±√{[9/16]}
  • tanθ = ±[3/4] Remember the mnemonic ASTC. This tells which quadrant tangent values will be positive. Tangent is only positive in quadrants I and III.
tanθ = − [3/4] because θ is in quadrant II
Verify the following identity: (tan2θ + 1)(cos2x - 1) = - tan2x
  • Try to get the left hand side to look like the right hand side because it is the more complicated side
  • (sec2θ)( - sin2x) = - tan2x by the Pythagorean Identities of sin2θ + cos2θ = 1 and 1 + tan2θ = sec2θ
  • [1/(cos2x)] · - sin2x = - tan2x by the Reciprocal Identity
  • [( − sin2x)/(cos2x)] = - tan2x by multiplying
  • ( [( − sinx)/cosx] )2 = - tan2x by the Rule of Exponents
- tan2x = - tan2x by the Quotient Identity
Verify the following identity: secθ + tanθ = [(cosθ)/(1 − sinθ)]
  • Try to get the right hand side to look like the left hand side bcause it is the more complicated side
  • secθ + tanθ = [(cosθ)/(1 − sinθ)] · [(1 + sinθ)/(1 + sinθ)], multiple by the conjugate
  • secθ + tanθ = [(cosθ+ cosθsinθ)/(1 − sin2θ)], multiplication of fractions
  • secθ + tanθ = [(cosθ+ cosθsinθ)/(cos2θ)], Pythagorean Identity sin2θ + cos2θ = 1
  • secθ + tanθ = [(cosθ)/(cos2θ)] + [(cosθsinθ)/(cos2θ)], Separate Fractions
  • secθ + tanθ = [1/(cosθ)] + [(sinθ)/(cosθ)], Simplifying
secθ + tanθ = secθ + tanθ, Reciprocal Identities
Verify the following identity: tanθ + cotθ = secθcscθ
  • Try to get the left hand side to look like the right hand side bcause it is the more complicated side
  • [(sinθ)/(cosθ)] + [(cosθ)/(sinθ)] = secθcscθ, Quotient Identity
  • [(sin2θ+ cos2θ)/(cosθsinθ)] = secθcscθ, Adding Fractions
  • [1/(cosθsinθ)] = secθcscθ, Pythagorean Identity
  • [1/(cosθ)] · [1/(sinθ)] = secθcscθ, Product of Fractions
secθcscθ = secθcscθ, Reciprocal Identities
Verify the following identity: [(cot2θ)/(cscθ)] = cscθ - sinθ
  • Try to get the left hand side to look like the right hand side bcause it is the more complicated side
  • [(csc2θ− 1)/(cscθ)] = cscθ - sinθ, Pythagorean Identity: csc2θ - 1 = cot2θ
  • [(csc2theta)/(cscθ)] - [1/(cscθ)] = cscθ - sinθ, Separate Fractions
  • cscθ - [1/(cscθ)] = cscθ - sinθ, Simplifying
cscθ - sinθ = cscθ - sinθ, Reciprocal Identity
Verify the following identity: cot2θ(sec2θ - 1) = 1
  • Try to get the left hand side to look like the right hand side bcause it is the more complicated side
  • cot2θtan2θ = 1, Pythagorean Identityq: tan2θ = sec2θ - 1
  • ( [(cos2θ)/(sin2θ)] ) ( [(sin2θ)/(cos2θ)] ) = 1, Reciprocal Identities
1 = 1, Simplifying
Verify the following identity: [(cot2θ)/(1 + cscθ)] = [(1 − sinθ)/(sinθ)]
  • [(csc2θ− 1)/(1 + cscθ)] = [(1 − sinθ)/(sinθ)], Pythagorean Identity on the left hand side
  • [((cscθ− 1)(cscθ+ 1))/(1 + cscθ)] = [(1 − sinθ)/(sinθ)], Factor the left hand side
  • cscθ - 1 = [(1 − sinθ)/(sinθ)], Simplify the left hand side
  • cscθ - 1 = [1/(sinθ)] - [(sinθ)/(sinθ)], Separate Fractions on the right hand side
cscθ - 1 = cscθ - 1, Reciprocal Identity on the right hand side

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Identity Tan(squared)x+1=Sec(squared)x

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Main Formulas 0:19
    • Companion to Pythagorean Identity
    • For Cotangents and Cosecants
    • How to Remember
  • Example 1: Prove the Identity 1:40
  • Example 2: Given Tan Find Sec 3:42
  • Example 3: Prove the Identity 7:45
  • Extra Example 1: Prove the Identity
  • Extra Example 2: Given Sec Find Tan

Transcription: Identity Tan(squared)x+1=Sec(squared)x

We are talking about the trigonometric identities in particular to the Pythagorean identities for (tan) and (cot) of (theta), and (sec) and (cosec).0000

Here we are being asked to prove the identity (cot)2 +1 = (cosec)2.0012

The trick there is to remember the original Pythagorean identity.0020

Let me write that down to start with, sin2x + cos2x = 1, that is the original Pythagorean identity.0024

That should be very familiar to you because you use that all the time in trigonometry.0035

We will start with that and I know I’m trying to find (cot) and (cosec) in this somehow.0043

What I’m going to do is divide both sides by sin2x.0049

Divide everything here by sin2x and that gives me 1.0063

Now (cos)2/(sin)2 that is (cos)/(sin)2, that is (cot)2x.0072

1/(sin)2 that is 1/(sin), that is (cosec) by definition.0081

There you have it, that is a pretty quick one.0090

We start with the original Pythagorean identity and we just divide both sides by (sin)2x to make it look exactly like what we are asked for.0093

I can rearrange terms here and I can get (cot)2x + 1 is equal to (cosec)2x.0104

This just comes back to knowing the original Pythagorean identity, if you remember that original Pythagorean identity.0119

And if you remember the definitions of sec, tan, cosec, and cot, then you can pretty quickly derive the new one cot2 + 1 = cosec2.0125

You might not even really need to memorize that one as long as you know the others vey well because you can work it out quickly.0136

Hi we are given (sec) of an angle here, 13/12 and we are given the angle in terms of degrees is between 270 and 360 and we are asked to find the tan(theta).0000

This is pretty clearly asking us to use the Pythagorean identity 10 2(theta) +1= sec2(theta).0012

Sec2(theta) is we plug in 13/12, that is 13/122 and that is 169/144.0023

102(theta) if we subtract 1 from both sides is 169/144 – 1 which is 169-144/144 which is 25/144.0038

Tan(theta) if we take the square root of both sides is + or – the square root of 25/144, which is + or -, those are both perfect squares so it comes out neatly 5/12.0067

Now the question is, whether it is the positive one or the negative one and we need to figure that out for the problem.0088

There is extra information given in the problem that we have not use yet.0094

We have not use the fact that (theta) is between 270 and 360 degrees .0098

Let me draw our (theta) would be approximately, here is my unit circle, in degrees that is 0, 90, 180, 270, and 360.0104

(theta) is between 270 and 360, (theta) is down there somewhere, so (theta) is an angle around there.0123

If you remember all students take calculus, that tells you which of these basic functions are positive and negative in each quadrants.0133

In the first quadrant they are all positive, in the second quadrant only (sin) is, third quadrant only (tan) is, fourth quadrant only (cos) is.0145

Our (theta) is in the fourth quadrant, let me write that as (theta) is in quadrant 4.0156

Tan(theta) only cos is positive there, tan is not, so tan(theta) is negative.0177

When we are choosing between these positive and negative square root here, we know we have to pick the negative one, -5/12.0188

There are really two steps to figuring out how to do this problem, one is to remember the Pythagorean identity tan2(theta) + 1=sec2(theta).0210

Then we take the value of sec(theta) that we are given, we plug it in, we work it down through and we try to solve for tan(theta) and we end up taking the square root.0221

We get plus or minus in there and we do not know what we want if positive or negative.0232

The second step there was to use the information about what quadrant (theta) is in and once we know that (theta) is in quadrant 4, we know that its tan it has to be negative.0237

That is from the all student take calculus rule and tan(theta) is negative, we know we need to take the negative square root.0253

That is how we get tan(theta) is equal to -5/12.0260

That is our lesson on the Pythagorean identity tan2(theta) + 1= sec2(theta).0265

These are the trigonometry lectures for www.educator.com.0271

OK, welcome back to the trigonometry lectures on educator.com, and today we're talking about the identity, tan2(x)+1=sec2(x).0000

You really want to think about this as a kind of companion identity to the main Pythagorean identity, which is that sin2(x)+cos2(x)=1.0010

Hopefully, you've really memorized this identity, sin2(x)+cos2(x)=1, that's one that comes up everywhere in trigonometry.0022

Then sort of the companion Pythagorean identity to that for secants and tangents is tan2(x)+1=sec2(x).0034

There's another related identity, essentially just the same for cotangents and cosecants, is that cot2(x)+1=csc2(x).0045

You may wonder how you'll remember this identities.0058

For example, how do you remember whether it's tan2+1=sec2, or maybe it's the other way around, sec2+1=tan2.0062

Well, really the answer to that lies in the exercises that we're about to do.0072

We'll show you how to derive those identities from the original one sin2+cos2=1.0077

As long as you can remember sin2+cos2=1, you'll learn how you can use that to figure out the others and you really won't have to work hard to remember this new Pythagorean identities.0085

Let's jump right into that.0099

The first example here is to prove the identity tan2(x)+1=sec2(x).0102

The trick there is to remember the original Pythagorean identity, which is cos2(x)+sin2(x)=1, that's the original Pythagorean identity.0109

That's one that you really should memorize and remember throughout all your work with trigonometry.0127

What you do to manipulate this into the new identity, is you just divide both sides by cos2(x).0136

On the left, we get cos2(x)/cos2(x) plus, let me write it as (sin(x)/cos(x)2=1/cos2(x).0152

Then of course, cos2(x)/cos2(x) is just 1, sin/cos, remember that's the definition of tangent, so this is tan2(x), 1/cos, that's the definition of sec(x), so sec2(x).0173

You can just rearrange this into tan2(x)+1=sec2(x).0191

That shows that this new identity is just a straight consequence of the original Pythagorean identity, so really if you can remember that Pythagorean identity, you can pretty quickly work out this new Pythagorean identity for tangents and cotangents.0203

Let's try using the Pythagorean identity for tangents and cotangents.0224

In this problem, we're given the tan(θ)=-4.21, and θ is between π/2 and π, we want to find secθ.0230

Remembering the Pythagorean identity, tan2(θ)+1=sec2(θ).0238

If we plug in what we're given here, tan2(θ), that's (-4.212)+1=sec2(θ).0246

Well, 4.212, that's something I'm going to workout on my calculator, is 17.7241.0259

So, 17.7241+1=sec2(θ), that's 18.7241=sec2(θ).0274

If I take the square root of both sides, I get sec(θ) is equal to plus or minus the square root of 18.7241.0290

Again, I'm going to do that square root on the calculator, and I get approximately 4.33.0303

The question here is whether we want the positive or negative square root, and that's where we use the other piece of information given in the problem.0323

Θ is in the second quadrant here, θ is between π/2 and π, so θ is somewhere over here.0332

Now, sec(θ) remember, is 1/cos(θ).0348

In the second quadrant, if you remember All Students Take Calculus.0353

In the second quadrant, sine is positive, but cosine is not, cosine is negative.0358

That means, sec(θ) is negative.0365

θ is in quadrant 2, cos(θ) is negative, sec(θ), because that's 1/(θ), is negative.0370

So, sec(θ), we want to take the negative square root there, and we get an approximate value of -4.33.0394

That negative is very important.0411

The key to that problem is first of all invoking this Pythagorean identity, tan2+1=sec2.0415

That's very important to remember.0423

We plug in the value that we're given and then we work it through and we'll try to solve for sec(θ).0426

We get this plus or minus at the end because we don't know if we want the positive square root or the negative square root.0431

That's where we use the fact that θ is in the second quadrant.0437

Since θ is in the second quadrant, cos(θ) must be negative, sec(θ), remember, is just 1/cos(θ), must also be negative.0448

That's how we know we want the negative square root there, so we get -4.33.0458

In our next example, we're given trigonometric identity and we're asked to prove it, (csc(θ)-cot(θ))/(sec(θ)-1) = cot(θ).0467

I'm going to start with the left-hand side of this trigonometric identity and I'm going to manipulate it.0480

I'm going to try and manipulate it into the right-hand side.0490

The left-hand side which is (csc(θ)-cot(θ))/(sec(θ)-1).0496

Now, I'm going to do something clever here and it's based on this old principle of whenever you have an (a-b) in the denominator, try multiplying by the conjugate, (a+b)/(a+b).0511

If it's the other way around, if you have (a+b), you multiply by the conjugate (a-b)/(a-b).0524

The reason you do that is that it makes the denominator (a2-b2).0530

We're taking advantage of that old formula from algebra, the difference of squares formula, and a lot of times the (a2-b2) is something that it'll simplify into something nice.0538

That's what's going to happen in here.0549

We're going to multiply, since we have (sec(θ)-1 in the denominator, by sec(θ)+1.0552

What that turns into in the denominator is this (a2-b2) form, so sec2(θ)-1.0565

The numerator really doesn't have anything very good happening yet, (csc(θ)-cot(θ))×sec(θ)+1, nothing very good happening in the numerator yet.0577

In the denominator, we're going to take advantage of the fact that tan2(θ)+1=sec2(θ), that's the Pythagorean identity that we're studying today.0593

If you move that around, you get sec2(θ)-1=tan2(θ).0607

That converts the denominator into tan2(θ), so that's very useful.0614

In the numerator, we have csc(θ)-cot(θ) and sec(θ)+1.0623

I think now I'm going to translate everything into sines and cosines because those will be easier to understand than cosecants and cotangents.0628

So, cosecant, remember is 1/sin(θ), minus cotangent is cos(θ)/sin(θ), sec(θ)+1, well, sec(θ) is 1/cos(θ), and tan(θ), if we translate into sines over cosines, that's sin2(θ)/cos2(θ).0637

Now, I've got fractions divided by fractions, I think the easiest thing to do here is to bring the denominator, flip it up and multiply it by the numerator.0670

We get cos2(θ)/sin2(θ), that's flipping up the denominator, and then the old numerator is, if I combined the first two terms of the first one, it's (1-cos(θ))/sin(θ).0681

In the second one, we have 1/cos(θ)+1.0706

Now, I think what I'm going to do is I'm going to look at this cosine squared, write it as cos(θ)×cos(θ).0715

Then pull one of those cosines over and multiply it by this term, and try to simplify things a little bit that way.0726

That will leave me with just one cosine left, still have sin2 in the denominator, (1-cos(θ)/sin(θ)) times, now, cos(θ)×(1/cos(θ) is 1, plus 1×cos(θ).0736

Now, look, we have (1-cos(θ))×(1+cos(θ)).0767

We're going to use this pattern again, the (a-b)×(a+b)=a2-b2.0772

If cos(θ)/sin2(θ), now multiply (1-cos)×(1+cos) gives us (1-cos2(θ)) and sin(θ) in the denominator.0780

Now, we're going to use the other Pythagorean identity, the original one which are right over here, sin2(θ)+cos2(θ)=1, if you switched that around 1-cos2(θ)=sin2(θ).0798

I'll plug that in, cos(θ)/sin(θ), 1-cos2(θ), now translates into sin2(θ) all divided by sin(θ).0819

I forgot my squared there, that's from the line above.0836

Now, we have some cancellations, sin2(θ) cancel, we get cos(θ)/sin(θ), but that's cot(θ).0840

By definition of cotangent, cotangent is defined to be cos/sin.0852

That's exactly equal to the right-hand side of the identity.0858

When you're proving this trigonometric identities, you pick one side and you multiply it as much as you can.0864

You try to manipulate it into the other side.0870

It does takes some trial and error.0874

I worked this problem out ahead of time, I tried a couple of different things and I finally found a way that gets us through it relatively quickly.0876

It's not like you'll always know exactly which path to follow, it takes a little bit of experimentation.0882

There are some patterns that you see over and over again, and the two patterns that we really invoke strongly in this one are these algebra pattern where (a-b)×(a+b)=a2-b2.0888

Essentially, whenever you see an (a-b) or an (a+b), think about multiplying it by the conjugate, and then taking advantage of that pattern.0904

That's what really got us started on the first step here.0913

We had a sec(θ)-1, and I thought, okay, let's multiply that by sec(θ)+1, that gives us sec2-1.0917

The second pattern that we use there was to invoke these Pythagorean identities, tan2+1=sec2, and sin2+cos2=1.0927

We invoked that one here, converting sec2-1 into tan2.0942

Then later on, when he had another (a-b)×(a+b), it converted into 1-cos2 and that in turn, by the Pythagorean identity converted into sin2.0947

It's just a lot of manipulation but you can let your manipulation be kind of guided by these common algebraic identities and these common Pythagorean identities.0962

Then you just try to keep manipulating until you get to the other side of the equation.0973

We'll try some more examples later on.0978

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