For more information, please see full course syllabus of Trigonometry
For more information, please see full course syllabus of Trigonometry
Identity Tan(squared)x+1=Sec(squared)x
Main formulas:
 For any angle x for which the tangent and secant are defined, we have tan^{2} x + 1 = sec^{2} x.
 For any angle x for which the tangent and secant are defined, we have cot^{2} x + 1 = csc^{2} x.
Example 1:
Prove the identity tan^{2} x + 1 = sec^{2} x.Example 2:
Given that tanθ = − 4.21 and (π/2) < θ < π , find secθ .Example 3:
Prove the following trigonometric identity:

Example 4:
Prove the identity cot^{2} x + 1 = csc^{2} x.Example 5:
Given that secθ = (13/12) and 270^{° } < θ < 360^{° } , find tanθ .Identity Tan(squared)x+1=Sec(squared)x
 Use the Pythagorean Identity to solve for secθ: 1 + tan^{2}θ = sec^{2}θ
 1 + ( − 5.31)^{2} = sec^{2}θ ⇒ 1 + 28.1961 = sec^{2}θ ⇒ sec^{2}θ = 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.
 Use the Pythagorean Identity to solve for tanθ: 1 + tan^{2}θ = sec^{2}θ
 1 + tan^{2}θ = ( [13/5] )^{2} ⇒ 1 + tan^{2}θ = [169/25] ⇒ tan^{2}θ = [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.
 Use the Pythagorean Identity to solve for secθ: 1 + tan^{2}θ = sec^{2}θ
 1 + ( − 3.23)^{2} = sec^{2}θ ⇒ 1 + 10.4329 = sec^{2}θ ⇒ sec^{2}θ = 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.
 Use the Pythagorean Identity to solve for tanθ: 1 + tan^{2}θ = sec^{2}θ
 1 + tan^{2}θ = ( − [5/4] )^{2} ⇒ 1 + tan^{2}θ = [25/16] ⇒ tan^{2}θ = [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.
 Try to get the left hand side to look like the right hand side because it is the more complicated side
 (sec^{2}θ)(  sin^{2}x) =  tan^{2}x by the Pythagorean Identities of sin^{2}θ + cos^{2}θ = 1 and 1 + tan^{2}θ = sec^{2}θ
 [1/(cos^{2}x)] ·  sin^{2}x =  tan^{2}x by the Reciprocal Identity
 [( − sin^{2}x)/(cos^{2}x)] =  tan^{2}x by multiplying
 ( [( − sinx)/cosx] )^{2} =  tan^{2}x by the Rule of Exponents
 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 − sin^{2}θ)], multiplication of fractions
 secθ + tanθ = [(cosθ+ cosθsinθ)/(cos^{2}θ)], Pythagorean Identity sin^{2}θ + cos^{2}θ = 1
 secθ + tanθ = [(cosθ)/(cos^{2}θ)] + [(cosθsinθ)/(cos^{2}θ)], Separate Fractions
 secθ + tanθ = [1/(cosθ)] + [(sinθ)/(cosθ)], Simplifying
 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
 [(sin^{2}θ+ cos^{2}θ)/(cosθsinθ)] = secθcscθ, Adding Fractions
 [1/(cosθsinθ)] = secθcscθ, Pythagorean Identity
 [1/(cosθ)] · [1/(sinθ)] = secθcscθ, Product of Fractions
 Try to get the left hand side to look like the right hand side bcause it is the more complicated side
 [(csc^{2}θ− 1)/(cscθ)] = cscθ  sinθ, Pythagorean Identity: csc^{2}θ  1 = cot^{2}θ
 [(csc^{2}theta)/(cscθ)]  [1/(cscθ)] = cscθ  sinθ, Separate Fractions
 cscθ  [1/(cscθ)] = cscθ  sinθ, Simplifying
 Try to get the left hand side to look like the right hand side bcause it is the more complicated side
 cot^{2}θtan^{2}θ = 1, Pythagorean Identityq: tan^{2}θ = sec^{2}θ  1
 ( [(cos^{2}θ)/(sin^{2}θ)] ) ( [(sin^{2}θ)/(cos^{2}θ)] ) = 1, Reciprocal Identities
 [(csc^{2}θ− 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
*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 screencaptured 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
Trigonometry Online Course
I. Trigonometric Functions  

Angles  39:05  
Sine and Cosine Functions  43:16  
Sine and Cosine Values of Special Angles  33:05  
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D  52:03  
Tangent and Cotangent Functions  36:04  
Secant and Cosecant Functions  27:18  
Inverse Trigonometric Functions  32:58  
Computations of Inverse Trigonometric Functions  31:08  
II. Trigonometric Identities  
Pythagorean Identity  19:11  
Identity Tan(squared)x+1=Sec(squared)x  23:16  
Addition and Subtraction Formulas  52:52  
Double Angle Formulas  29:05  
HalfAngle Formulas  43:55  
III. Applications of Trigonometry  
Trigonometry in Right Angles  25:43  
Law of Sines  56:40  
Law of Cosines  49:05  
Finding the Area of a Triangle  27:37  
Word Problems and Applications of Trigonometry  34:25  
Vectors  46:42  
IV. Complex Numbers and Polar Coordinates  
Polar Coordinates  1:07:35  
Complex Numbers  35:59  
Polar Form of Complex Numbers  40:43  
DeMoivre's Theorem  57:37 
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, sin^{2}x + cos^{2}x = 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 Iâ€™m trying to find (cot) and (cosec) in this somehow.0043
What Iâ€™m going to do is divide both sides by sin^{2}x.0049
Divide everything here by sin^{2}x and that gives me 1.0063
Now (cos)^{2}/(sin)^{2} that is (cos)/(sin)^{2}, that is (cot)^{2}x.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)^{2}x to make it look exactly like what we are asked for.0093
I can rearrange terms here and I can get (cot)^{2}x + 1 is equal to (cosec)^{2}x.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 cot^{2} + 1 = cosec^{2}.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= sec^{2}(theta).0012
Sec^{2}(theta) is we plug in 13/12, that is 13/12^{2} and that is 169/144.0023
10^{2}(theta) if we subtract 1 from both sides is 169/144 â€“ 1 which is 169144/144 which is 25/144.0038
Tan(theta) if we take the square root of both sides is + or â€“ 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 tan^{2}(theta) + 1=sec^{2}(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 tan^{2}(theta) + 1= sec^{2}(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, tan^{2}(x)+1=sec^{2}(x).0000
You really want to think about this as a kind of companion identity to the main Pythagorean identity, which is that sin^{2}(x)+cos^{2}(x)=1.0010
Hopefully, you've really memorized this identity, sin^{2}(x)+cos^{2}(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 tan^{2}(x)+1=sec^{2}(x).0034
There's another related identity, essentially just the same for cotangents and cosecants, is that cot^{2}(x)+1=csc^{2}(x).0045
You may wonder how you'll remember this identities.0058
For example, how do you remember whether it's tan^{2}+1=sec^{2}, or maybe it's the other way around, sec^{2}+1=tan^{2}.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 sin^{2}+cos^{2}=1.0077
As long as you can remember sin^{2}+cos^{2}=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 tan^{2}(x)+1=sec^{2}(x).0102
The trick there is to remember the original Pythagorean identity, which is cos^{2}(x)+sin^{2}(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 cos^{2}(x).0136
On the left, we get cos^{2}(x)/cos^{2}(x) plus, let me write it as (sin(x)/cos(x)^{2}=1/cos^{2}(x).0152
Then of course, cos^{2}(x)/cos^{2}(x) is just 1, sin/cos, remember that's the definition of tangent, so this is tan^{2}(x), 1/cos, that's the definition of sec(x), so sec^{2}(x).0173
You can just rearrange this into tan^{2}(x)+1=sec^{2}(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, tan^{2}(θ)+1=sec^{2}(θ).0238
If we plug in what we're given here, tan^{2}(θ), that's (4.21^{2})+1=sec^{2}(θ).0246
Well, 4.21^{2}, that's something I'm going to workout on my calculator, is 17.7241.0259
So, 17.7241+1=sec^{2}(θ), that's 18.7241=sec^{2}(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, tan^{2}+1=sec^{2}.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 lefthand side of this trigonometric identity and I'm going to manipulate it.0480
I'm going to try and manipulate it into the righthand side.0490
The lefthand 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 (ab) 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 (ab)/(ab).0524
The reason you do that is that it makes the denominator (a^{2}b^{2}).0530
We're taking advantage of that old formula from algebra, the difference of squares formula, and a lot of times the (a^{2}b^{2}) 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 (a^{2}b^{2}) form, so sec^{2}(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 tan^{2}(θ)+1=sec^{2}(θ), that's the Pythagorean identity that we're studying today.0593
If you move that around, you get sec^{2}(θ)1=tan^{2}(θ).0607
That converts the denominator into tan^{2}(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 sin^{2}(θ)/cos^{2}(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 cos^{2}(θ)/sin^{2}(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 (1cos(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 sin^{2} in the denominator, (1cos(θ)/sin(θ)) times, now, cos(θ)×(1/cos(θ) is 1, plus 1×cos(θ).0736
Now, look, we have (1cos(θ))×(1+cos(θ)).0767
We're going to use this pattern again, the (ab)×(a+b)=a^{2}b^{2}.0772
If cos(θ)/sin^{2}(θ), now multiply (1cos)×(1+cos) gives us (1cos^{2}(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, sin^{2}(θ)+cos^{2}(θ)=1, if you switched that around 1cos^{2}(θ)=sin^{2}(θ).0798
I'll plug that in, cos(θ)/sin(θ), 1cos^{2}(θ), now translates into sin^{2}(θ) all divided by sin(θ).0819
I forgot my squared there, that's from the line above.0836
Now, we have some cancellations, sin^{2}(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 righthand 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 (ab)×(a+b)=a^{2}b^{2}.0888
Essentially, whenever you see an (ab) 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 sec^{2}1.0917
The second pattern that we use there was to invoke these Pythagorean identities, tan^{2}+1=sec^{2}, and sin^{2}+cos^{2}=1.0927
We invoked that one here, converting sec^{2}1 into tan^{2}.0942
Then later on, when he had another (ab)×(a+b), it converted into 1cos^{2} and that in turn, by the Pythagorean identity converted into sin^{2}.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
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(ÃŽÂ¸)+1 = sec^2(ÃŽÂ¸), but it also shows that sometimes there are simpler solutions, like e.g. this one:
LHS = (cscÃŽÂ¸cotÃŽÂ¸)/(secÃŽÂ¸1) = (1/sinÃŽÂ¸cosÃŽÂ¸/sinÃŽÂ¸)/(1/cosÃŽÂ¸1) =
= ((1cosÃŽÂ¸)/sinÃŽÂ¸)/((1cosÃŽÂ¸)/cosÃŽÂ¸) = (1/sinÃŽÂ¸)/(1/cosÃŽÂ¸) =
= cosÃŽÂ¸/sinÃŽÂ¸ = cotÃŽÂ¸ = 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.