Question

Verify the identity. Justify each step.

Answer 1

\[\frac{ \sec 0 }{ \csc 0-\cot 0 }-\frac{ \sec 0 }{ \csc 0+\cot 0 }=2\csc 0\]

Answer 2

@imsleeziboii can you help?

Answer 3

I honestly have no clue on this, i dont think im to this point in math yet. Sorry

Answer 4

its fine :/

Answer 5

@freckles ....Can you help me?!!!!!?

Answer 6

i guess the number is 0 not actually 0 here
it is some variable like theta

Answer 7

my first step would be to attempt to find a common denominator so i could combine the fractions

Answer 8

ok.........how do i do that?

Answer 9

\[\frac{a}{b}+\frac{c}{d} \\ =\frac{ad+cb}{bd}\]

Answer 10

ok sooo
\[\frac{ \sec0+\sec }{ \csc \cot }\]
???

Answer 11

no you need to combine the fractions like I have in my example above

Answer 12

\[\frac{\sec(x)(\csc(x)+\cot(x))-\sec(x)(\csc(x)-\cot(x))}{(\csc(x)-\cot(x))(\csc(x)+\cot(x))}\]

Answer 13

multiply stuff out now

Answer 14

oh dang

Answer 15

\[\frac{ (\sec)(\csc)+(\cot)-(\sec)(\csc)-(\cot) }{ (\csc-\cot)(\csc+\cot)}\]
?????

Answer 16

looks like you didn't use distributive property correctly in the numerator
and you also need to multiply the bottom out
also it is weird to write sec and csc and cot by themselves
should by sec(x) and csc(x) and cot(x)

Answer 17

uggghhh i suck at these

Answer 18

are you familiar with distributive property?

Answer 19

a(b+c)=ab+ac

Answer 20

yea

Answer 21

so you see that when you applied distributive property
you only half did it correctly since you only distributed to half the terms in the ( )
I'm saying you did this a(b+c)=ab+c which is incorrect
you need to multiply all the terms in ( ) by a

Answer 22

......you lost me

Answer 23

ok so you don't know distributive property is what you are saying?

Answer 24

a(b+c) means we need to multiply all the terms in ( ) by a
so
a(b+c) is the same as ab+ac

Answer 25

no i do, but what you just said confused me alot

Answer 26

you only multiply the first term in the ( ) by a
but you had two terms

Answer 27

sec(x)(cot(x)+csc(x))
is in the form
a(b+c)
a=sec(x)
b=cot(x)
c=csc(x)

a(b+c)+ab+ac
not ab+c

Answer 28

a(b+c)=ab+ac*

Answer 29

ok i get what your saying now, but how do i multiply them?

Answer 30

by using the property have been talking about

Answer 31

you try
sec(x)(csc(x)+cot(x))=?

Answer 32

you say you know how to use a(b+c)=ab+ac

Answer 33

that is all I'm asking you to do here

Answer 34

yes with funny letters that are crazy and no one should have to know!

Answer 35

what funny letters?

Answer 36

csc cot tan sin ect.

Answer 37

\[\color{red}a(\color{blue}b+\color{green}c)=\color{red}a\color{blue}b+\color{red}a \color{green}c \\ \text{ and we want \to apply this here } \\ \color{red}{\sec(x)}(\color{blue}{\csc(x)}+\color{green}{\cot(x)})=?\]

Answer 38

sec(x) is a number just like a is
csc(x) is a number just like b is
cot(x) is a number just like c is

Answer 39

what numbers?!

Answer 40

what do you mean what numbers ?

Answer 41

a,b,c are numbers just like sec(x),csc(x), and cot(x)

Answer 42

sec by itself is not a number
but sec(x) is a number

Answer 43

thats messed up

Answer 44

ok I will do the first one for you...but you have to do the next one by yourself
that is I will do:
\[\sec(x)(\csc(x)+\cot(x)) \text{ and you will do } -\sec(x)(\csc(x)-\cot(x))\]

\[\sec(x)(\csc(x)+\cot(x)) \\ =\sec(x) \csc(x)+\sec(x)\cot(x) \text{ by distributive property }\]

you do the other part the -sec(x)(csc(x)-cot(x)) part

Answer 45

\[-\sec(x)(\csc(x)-\cot(x))\]
\[= -\sec(x)(\csc(x)-\sec(x)\cot(x)\]
???

Answer 46

almost
it looks like you forgot that a - times - is +

Answer 47

\[\sec(x)(\csc(x)-\sec(x)\cot(x)\]

or

\[\sec(x)(\csc(x)+\sec(x)\cot(x))\]

Answer 48

did you mean to put the extra parenthesis ?
\[-\sec(x)(\csc(x)-\cot(x)) \\ =-\sec(x)\csc(x)--\sec(x)\cot(x) \\= -\sec(x)\csc(x)+\sec(x)\cot(x)\]

Answer 49

\[\frac{\sec(x)(\csc(x)+\cot(x))-\sec(x)(\csc(x)-\cot(x))}{(\csc(x)-\cot(x))(\csc(x)+\cot(x))} \\ =\frac{\sec(x)\csc(x)+\sec(x)\cot(x)-\sec(x)\csc(x)+\sec(x)\cot(x)}{(\csc(x)-\cot(x))(\csc(x)+\cot(x))}\]
ok you should combine like terms on top
and you still need to multiply the bottom out
brb

Answer 50

you can post and i will be right back to check

Answer 51

ok, actually i have to brb too i will as soon as i get back tho

Answer 52

\[(\csc(x)-\cot(x))(\csc(x)+\cot(x))\]
\[(\csc(x)+Cot(x))\]

Answer 53

well you have (a-b)(a+b)
which is a^2-b^2
not a+b

Answer 54

(a-b)(a+b)
a(a+b)-b(a+b)
aa+ab-ba-bb
a^2+ab-ab-b^2
a^2+0-b^2
a^2-b^2

so (a-b)(a+b)=a^2-b^2
you just need to multiply the first terms and last terms since you are multiplying conjugates

Answer 55

so
\[\csc^2-\cot^2\]

Answer 56

don't forget that csc and cot are meaningless with the of angle part
should be csc^2(x)-cot^2(x)

Answer 57

did you see any like terms in our numerator

Answer 58

\[\frac{\sec(x)(\csc(x)+\cot(x))-\sec(x)(\csc(x)-\cot(x))}{(\csc(x)-\cot(x))(\csc(x)+\cot(x))} \\ =\frac{\sec(x)\csc(x)+\sec(x)\cot(x)-\sec(x)\csc(x)+\sec(x)\cot(x)}{(\csc(x)-\cot(x))(\csc(x)+\cot(x))} \\ =\frac{\sec(x) \csc(x)+\sec(x)\cot(x)-\sec(x)\csc(x)+\sec(x)\cot(x)}{\csc^2(x)-\cot^2(x)}\]
look for like terms in numerator
and think about using a Pythagorean identity for the denominator

Answer 59

\[\sec^2(x)+\cot^2(x)-\csc^2(x)\]

Answer 60

what happened

Answer 61

do you see the like terms on top? group them together
sec(x)csc(x)-sec(x)csc(x)+sec(x)cot(x)+sec(x)cot(x)
can you finish simplifying top from here

also did you try to use a Pythagorean identity for the denominator ?

Answer 62

was trying to work with the top part before i got to the bottom part so i wouldn't get as confused

Answer 63

do you know how to do the following:
5-5=?
x-x=?
2x-2x=?
-5-(-5)=?
2xy-2xy=?
f(x)-f(x)=?
f(x)g(x)-f(x)g(x)=?
anything-itself=?
so
sec(x)csc(x)-sec(x)csc(x)=?

Answer 64

somewhat

Answer 65

if i give you x of something and I take away x of something what are you left with ?

Answer 66

5-5=0
x-x=0
2x-2x=0
-5-(-5)=0
2xy-2xy=0
f(x)-f(x)=0
f(x)g(x)-f(x)g(x)=0
twinkle little star-twinkle little star=0
anything-itself=0

Answer 67

zero

Answer 68

sec(x)csc(x)-sec(x)csc(x)=0

Answer 69

do you know how to do sec(x)cot(x)+sec(x)cot(x)=?

Answer 70

doesn't that equal 1?

Answer 71

anything added to itself is just double that thing

examples:
5+5=2(5) or 10
x+x=2x
f(x)+f(x)=2f(x)
f(x)g(x)+f(x)g(x)=2f(x)g(x)
xy+xy=2xy
twinkle little star+twinkle little star=2(twinkle little star)

Answer 72

......so 2

Answer 73

2what

Answer 74

2sec cot?

Answer 75

almost
sec and cot again are meaningless
you mean 2sec(x)cot(x)

Answer 76

\[\frac{\sec(x)(\csc(x)+\cot(x))-\sec(x)(\csc(x)-\cot(x))}{(\csc(x)-\cot(x))(\csc(x)+\cot(x))} \\ =\frac{\sec(x)\csc(x)+\sec(x)\cot(x)-\sec(x)\csc(x)+\sec(x)\cot(x)}{(\csc(x)-\cot(x))(\csc(x)+\cot(x))} \\ =\frac{\sec(x) \csc(x)+\sec(x)\cot(x)-\sec(x)\csc(x)+\sec(x)\cot(x)}{\csc^2(x)-\cot^2(x)} \\ =\frac{2 \sec(x)\cot(x)}{\csc^2(x)-\cot^2(x)}\]

Answer 77

can you apply a Pythagorean identity to the denominator ?

Answer 78

\[\csc^2=\cot(x)-3\]
is that right for that one?

Answer 79

please tell me that isn't a Pythagorean identity you have written down

Answer 80

...well im guessing thats wrong

Answer 81

\[\sin^2(x)+\cos^2(x)=1 \\ \tan^2(x)+1=\sec^2(x) \\ 1+\cot^2(x)=\csc^2(x)\]
these are Pythagorean identities

the thing you wrote isn't even an identity

Answer 82

THIS IS IMPOSSIBLE

Answer 83

no it isn't

Answer 84

like the last equation should have been the first equation your eyes went to
if you wanted to apply a Pythagorean identity to the denominator in our problem
why? because it involves cot^2(x) and csc^2(x) just like our denominator does

Answer 85

\[1+\cot^2(x)=\csc^2(x) \text{ use this for your denominator }\]

Answer 86

using that what is the simplified expression for the denominator

Answer 87

\[\csc^2(x)\]?

Answer 88

can you showed me how you got that please

Answer 89

if I told you 1+a^2=b^2
then what would be the value of b^2-a^2 ?

Answer 90

2?

Answer 91

:(

Answer 92

1+a^2=b^2 pretend this is an identity
then if I subtract a^2 on both sides
1=b^2-a^2 (then this is also an identity )

if 1+a^2=b^2 then b^2-a^2=1

Answer 93

ooh

Answer 94

so if 1+cot^2(x)=csc^2(x)
then csc^2(x)-cot^2(x) equals?

Answer 95

1!

Answer 96

yes!

Answer 97

YAY!!!!!!

Answer 98

\[\frac{\sec(x)(\csc(x)+\cot(x))-\sec(x)(\csc(x)-\cot(x))}{(\csc(x)-\cot(x))(\csc(x)+\cot(x))} \\ =\frac{\sec(x)\csc(x)+\sec(x)\cot(x)-\sec(x)\csc(x)+\sec(x)\cot(x)}{(\csc(x)-\cot(x))(\csc(x)+\cot(x))} \\ =\frac{\sec(x) \csc(x)+\sec(x)\cot(x)-\sec(x)\csc(x)+\sec(x)\cot(x)}{\csc^2(x)-\cot^2(x)} \\ =\frac{2 \sec(x)\cot(x)}{\csc^2(x)-\cot^2(x)} \\ =\frac{2 \sec(x) \cot(x)}{1} \\ =2 \sec(x) \cot(x)\]