Question

The expression (secx + tanx)2 is the same as _____.

A. 1 + 2cscx
B. sec2x + 2cscx + tan2x
C. sec2x + tan2x
D. 1 + 2tan2x + 2secx tanx

Answer 1

\[\Large\rm (\sec x + \tan x)^2\]Remember how to expand out a squared expression like this? :)\[\Large\rm (a+b)^2=a^2+2ab+b^2\]

Answer 2

Hmm so judging by the options, you'll have to simplify a lil bit after expanding...

Answer 3

thank you

Answer 4

If you're able to get past that first part,
here is a hint to help you finish it up.
Remember your Pythagorean identity for tangent,\[\Large\rm \color{red}{\sec^2x=1+\tan^2x}\]

Answer 5

pythagorean identity for tan?

Answer 6

These are the Pythagorean Identities:\[\Large\rm \sin^2x+\cos^2x=1\]\[\Large\rm 1+\tan^2x=\sec^2x\]\[\Large\rm 1+\cot^2x=\csc^2x\]

Answer 7

After you expand everything out, you can replace the sec^2x(that you'll end up with) with 1+tan^2x.

Answer 8

so i would get 1+tan^2x=1+tan^2x? then what?

Answer 9

i think i did something wrong there

Answer 10

could you please walk me through the whole thing?

Answer 11

oh wait i get it!

Answer 12

sorry just got off a flight and im trying to work while i wait for the next one

Answer 13

anyways thank you very much!

Answer 14

Think you got it? c: cool

Answer 15

definitely :) there is another question i was wondering if you could help me with though

Answer 16

Sure

Answer 17

Which of the following is not an identity?

A. secx cscx(tanx + cotx) = sec2x + csc2x
B. sin2x + tan2x + cos2x = sec2x
C. 2sin2x - sinx = 1
D. 2cos2x - 1 = 1 - 2sin2x

Answer 18

I'm not so great with identities

Answer 19

Are any of those angles (2x) or are they all squares?
It's kind of hard to tell the difference when you copy paste :P heh

Answer 20

For example is option C:\[\Large\rm 2\sin^2x-\sin x=1, \qquad\text{or}\qquad 2\sin(2x)-\sin x=1\]

Answer 21

they're all squares, sorry

Answer 22

Let's look at option B a moment:
\[\Large\rm \sin^2x + \tan^2x + \cos^2x = \sec^2x\]Let's move things around a lil bit,\[\Large\rm \color{royalblue}{\sin^2x + \cos^2x} + \tan^2x = \sec^2x\]

Answer 23

Hmm do you see an identity we can apply to the blue part?
Gotta remember those Pythagorean Identities!! :)
Scroll back up the page if you forgot them.

Answer 24

\[\sin^2x+\cos^2x=1\]

Answer 25

and then 1+tan^2x=sec^2x

Answer 26

Ok great! So option B is not what we're looking for.

Answer 27

alrighty, on to the next one

Answer 28

Option D:\[\Large\rm 2\cos^2x - 1 \quad=\quad 1 - 2\sin^2x\]Hmm let's see...

Answer 29

Let's use our Pythagorean Identity to substitute something in for cos^2x.\[\Large\rm \sin^2x+\cos^2x=1\qquad\to\qquad \color{orangered}{\cos^2x=1-\sin^2x}\]
We'll use that to replace our cos^2x,\[\Large\rm 2(\color{orangered}{\cos^2x}) - 1 \quad=\quad 1 - 2\sin^2x\]

Answer 30

Understand what I did with the Pythagorean Identity? I solved it for cos^2x.

Answer 31

yes :)

Answer 32

Plugging in gives us,
\[\Large\rm 2(\color{orangered}{1-\sin^2x}) - 1 \quad=\quad 1 - 2\sin^2x\]

Hmm will it simplify to match the other side or no?

Answer 33

yes it will

Answer 34

i think

Answer 35

Ok cool c: So we don't want option D!

Answer 36

so it will be c or a

Answer 37

This is a tough problem, it's really tedious, and requires you to know a lot of your identities.

Answer 38

Umm ok let's check out option A:\[\Large\rm \sec x \csc x (\tan x + \cot x) = \sec^2x + \csc^2x\]

Answer 39

In order to work with this one, I think we're going to want to convert everything to sines and cosines.

Answer 40

alright

Answer 41

Remember how to write secx, cscx, tanx and cotx in terms of sines and cosines?

Answer 42

no :(

Answer 43

You... <.<

Answer 44

Hahaha sorry, this is my math prep for precalc next year, i barely know anything

Answer 45

\[\Large\rm \sec x =\frac{1}{\cos x},\qquad \csc x=\frac{1}{\sin x}\]
\[\Large\rm \tan x=\frac{\sin x}{\cos x},\qquad \cot x=\frac{\cos x}{\sin x}\]

Answer 46

\[\Large\rm \sec x \csc x (\tan x + \cot x) = \sec^2x + \csc^2x\]
Plugging all those identities in gives us:\[\Large\rm \frac{1}{\sin x \cos x} \left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right) = \sec^2x + \csc^2x\]Ok? :o

Answer 47

Okay, that makes sense

Answer 48

We'll distribute the ugly fraction in front, to each term in the brackets.

Answer 49

\[\Large\rm \frac{\sin x}{\sin x \cos x\cos x} + \frac{\cos x}{\sin x\sin x\cos x} = \sec^2x + \csc^2x\]

Answer 50

Understand what I did there? :o I gave them each a 1/(sin x cos x)

Answer 51

yeah

Answer 52

Looks like we have some nice cancellations.
\[\Large\rm \frac{\cancel{\sin x}}{\cancel{\sin x} \cos x\cos x} + \frac{\cancel{\cos x}}{\sin x\sin x\cancel{\cos x}} = \sec^2x + \csc^2x\]

Answer 53

\[\Large\rm \frac{1}{\cos^2x}+\frac{1}{\sin^2x}=\sec^2x+\csc^2x\]

Answer 54

Is there a way we can write these back in terms of secant and cosecant?
Look back at the 4 identities I posted before! :O

Answer 55

it would be secx and cscx?

Answer 56

Lemme write the squares like this so maybe it's easier to see what's going on,\[\Large\rm \left(\frac{1}{\cos x}\right)^2+\left(\frac{1}{\sin x}\right)^2=\sec^2x+\csc^2x\]

Answer 57

We had an identity for 1/cos x, yes?

Answer 58

yes

Answer 59

\[\Large\rm \left(\color{royalblue}{\frac{1}{\cos x}}\right)^2+\left(\color{royalblue}{\frac{1}{\sin x}}\right)^2=\sec^2x+\csc^2x\]Plugging those identities in gives us,\[\Large\rm \left(\color{royalblue}{\sec x}\right)^2+\left(\color{royalblue}{\csc x}\right)^2=\sec^2x+\csc^2x\]

Answer 60

So is this one true? :O

Answer 61

i think so yes

Answer 62

Ok good! So it looks like we've narrowed it down to C!
I didn't mean to check every correct one first, that was on accident lol

Answer 63

hahaha it's fine :) it helped me understand the process! thank you so much!

Answer 64

yay team \c:/

Answer 65

woohoo :)