Question

can some check
cscθ secθ/csc^2θ+sec^2θ
csc^2x=1/sin^2x
x=1/cosx
sec^2x=x1/cosx
(1/sin^2x)+(1/cos^2x)
how do i simplfy?

Answer 1

@Nnesha

Answer 2

plz someone

Answer 3

is it one question or 4 ? :o

Answer 4

1

Answer 5

i showed my work

Answer 6

i wanted to know if it is correct

Answer 7

alright let me check it
is this ur question \[\large\rm \frac{ \csc \theta \sec \theta }{ \csc^2 \theta } =\sec^2 \theta \]

Answer 8

ohh okay wait a sec i'll check

Answer 9

k

Answer 10

i don't get what you mean by this `x=1/cosx `
but looks like you're working on the denominator \[\large\rm \frac{ \csc \theta \sec \theta }{ \csc^2 \theta + \sec^2 \theta } \rightarrow \frac{ \csc \theta \sec \theta }{ \frac{ 1 }{ \sin^2\theta }+\frac{ 1 }{ \cos^2 \theta }}\]

Answer 11

yes the denominator

Answer 12

alright good so far now first lets deal with the denominator
\[\large\rm \frac{ \csc \theta \sec \theta }{ \csc^2 \theta + \sec^2 \theta } \rightarrow \frac{ \csc \theta \sec \theta }{ \color{Red}{\frac{ 1 }{ \sin^2\theta }+\frac{ 1 }{ \cos^2 \theta } }}\] find common denominator

Answer 13

cos

Answer 14

\[\large\rm \frac{ 1 }{ \sin^2\theta }+\frac{ 1 }{ \cos^2 \theta }\]
hmm no both denominators are different so we should multiply them

Answer 15

csc2+sec2

Answer 16

here is an example
when we find common denominators \[\large\rm \frac{ 1 }{ x }+\frac{1}{y}\]
both denominators aren't the same so we should multiply them


\[\frac{ y +x}{ xy }\]
and when we find common denominator
we should multiply numerator of first fraction by the denominator of 2nd fraction
multiply numerator of 2nd fraction by the denominator of first fraction

Answer 17

no we changed csc^2+sec^2 to 1/sin^2theta +1/cos^2

Answer 18

\[\large\rm \frac{ 1 }{ \sin^2\theta }+\frac{ 1 }{ \cos^2 \theta }\]
what's common denomiantor ?

Answer 19

2

Answer 20

hmm there isn't any 2
let's
say sin^2 theta = a
and cos^2 theta = b

\[\large\rm \frac{ 1 }{a }+\frac{ 1 }{b}\]
what's the common denominator ?

Answer 21

cos

Answer 22

not just cos hmm look at the example i gave you

Answer 23

srry sin would be considered

Answer 24

yes both denominators aren't the same so we should multiply them and btw it's sin^2 and cos^2 not just sin cos

Answer 25

\[\frac{ ??? +??? }{ \sin^2 \theta \cos^2\theta }\] what would you get at the numerator ?

Answer 26

it would be csctheta sectheta

Answer 27

hmm no that's the numerator
but i was talking about the numerator of bottom fraction
\[\large\rm \frac{ \csc \theta \sec \theta }{ \color{Red}{\frac{ 1 }{ \sin^2\theta }+\frac{ 1 }{ \cos^2 \theta } }}\rightarrow \frac{ csc \theta sec \theta }{\frac{???+??}{sin^2 \theta +cos^2 \theta }}\]

Answer 28

1/sinx 1/cos

Answer 29

here is an example
when we find common denominators \[\large\rm \frac{ 1 }{ x }+\frac{1}{y}\]
both denominators aren't the same so we should multiply them


\[\frac{ y +x}{ xy }\]
and when we find common denominator
we should multiply numerator of first fraction by the denominator of 2nd fraction
multiply numerator of 2nd fraction by the denominator of first fraction

Answer 30

` multiply numerator of first fraction by the denominator of 2nd fraction`
`multiply numerator of 2nd fraction by the denominator of first fraction `

Answer 31

so it would be cos theta and sin theta

Answer 32

hmm not just cos it's cos^2

Answer 33

ok

Answer 34

don't forget the sign hmm no that's the numerator
\[\huge\rm \ \rm \frac{ csc \theta sec \theta }{\frac{cos^2 \theta +sin^2 theta }{sin^2 \theta *cos^2 \theta }}\]
i made a typo earlier it's sin^2 times cos^2 not +

Answer 35

now we can rewrite csc and sec in terms of cos sin \[\huge\rm \frac{ \frac{ 1 }{ \sin \theta } *\frac{1}{\cos^2\theta }}{ \frac{ \cos^2 \theta +\sin^2 \theta }{\sin^2 \theta *\cos^2 \theta}}\]

Answer 36

tr to simplify that
remember the special identity \[\sin^2 \theta +\cos^2 \theta =1\]tr to simplify this
remember the special identity \[\sin^2 \theta +\cos^2 \theta =1\]

Answer 37

try*

Answer 38

one sec

Answer 39

sure take u time :=)

Answer 40

ur**

Answer 41

sintheta

Answer 42

wait there is a typo

Answer 43

nnesha could you please help me on the question I tagged you in if you can?

Answer 44

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
now we can rewrite csc and sec in terms of cos sin \[\huge\rm \frac{ \frac{ 1 }{ \sin \theta } *\frac{1}{\cos^2\theta }}{ \frac{ \cos^2 \theta +\sin^2 \theta }{\sin^2 \theta *\cos^2 \theta}}\]
\(\color{blue}{\text{End of Quote}}\)
it's\[\huge\rm \frac{ \frac{ 1 }{ \sin \theta } *\frac{1}{\color{Red}{\cos\theta} }}{ \frac{ \cos^2 \theta +\sin^2 \theta }{\sin^2 \theta *\cos^2 \theta}}\]
cos theta

Answer 45

no it 's not sin
please show the steps how did you get that

Answer 46

so cos theta is the answer

Answer 47

how did you get tthat ?

Answer 48

ok so sin theta was right

Answer 49

how did you get that please show ur work

Answer 50

i guessed it

Answer 51

after doing all this work now we should guess the answer ???

Answer 52

srry

Answer 53

i also have notes

Answer 54

\[\huge\rm \frac{ \frac{ 1 }{ \sin \theta } *\frac{1}{\cos\theta }}{ \frac{\color{Red}{ \cos^2 \theta +\sin^2 \theta} }{\sin^2 \theta *\cos^2 \theta}}\]
cos^2 x+sin^2 x= ??

Answer 55

1

Answer 56

right \[\huge\rm \frac{ \frac{ 1 }{ \sin \theta } *\frac{1}{\cos\theta }}{ \frac{\color{Red}{ 1} }{\sin^2 \theta *\cos^2 \theta}}\]
now change division to multiplication

Answer 57

\[\frac{ 1 }{ \sin \theta }*\frac{1}{\cos \theta} = \frac{1}{ \sin \theta*\cos \theta}\]

Answer 58

thats my answer

Answer 59

no.

Answer 60

\[\huge\rm \frac{ \frac{ 1 }{ \sin \theta*cos \theta } }{ \frac{\color{Red}{ 1} }{\sin^2 \theta *\cos^2 \theta}}\]
now change division to multiplication

Answer 61

csc3(θ)sec3(θ)

Answer 62

hmm here is an example \[\large\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d } }=\frac{ a }{ b }*\frac{ d }{ c }\] multiply top fraction by the `reciprocal ` of the bottom fraction

Answer 63

sine

Answer 64

show me how did you get that ?

Answer 65

i did 1/sintheta*costheta

Answer 66

tag my name ilost ur post

Answer 67

nvm found it

Answer 68

@Nnesha

Answer 69

alright what's the reciprocal of bottom fraction ??

Answer 70

Would it be cos sin

Answer 71

I suck at math

Answer 72

\[\huge\rm \frac{ \color{Red}{\frac{ 1 }{ \sin \theta*cos \theta } }}{ \color{blue}{\frac{ 1 }{\sin^2 \theta *\cos^2 \theta}}}\]

what's the reciprocal of bottom fraction ?

Answer 73

this is like pre-calculus but we are dealing with algebra stuff practice will make it easy

Answer 74

here is an example reciprocal of x/y is y/x

Answer 75

Cos theta/sin theta

Answer 76

hmm no \[\frac{ 1 }{ \sin^2 \theta* \cos^2 \theta }\] just flip the fraction !

Answer 77

Sin2theta cos2theta/1

Answer 78

\[\frac{ 1 }{ \cos \theta*\sin \theta } *\frac{\sin^2 \theta \cos^2 \theta}{1}\]
now simplify please

Answer 79

Cos theta sin theta

Answer 80

correct

Answer 81

So that's my answer

Answer 82

yep.

Answer 83

Thank you for the help I really do appreciate it:)

Answer 84

np