Question

Help with: (sec^2x)(csc^2x) - sec^2x - csc^2x + 1?

Answer 1

I'm not really sure how you can factor this.

Answer 2

sec(2x)csc(2x)−sec(2x)−csc(2x+1)

Answer 3

rewrite all sec and csc in terms of cos and sin

Answer 4

what's the reciprocal of sec ?

Answer 5

1/cos

Answer 6

correct so since it's sec^2x it would be 1/cos^2x
what aabout csc^2x ?

Answer 7

\[\frac{ 1 }{ \sin^2 x }\]

Answer 8

So:

\[\frac{ 1 }{ \cos^2 x } x \frac{ 1 }{ \sin^2 x } - \frac{ 1 }{ \cos^2 x } - \frac{ 1 }{ \sin^2 x }\]

Answer 9

correct \[\rm \frac{ 1 }{ \cos^2x}*\frac{1}{\sin^2x}-\frac{1}{\cos^2x}-\frac{1}{\sin^2x}+1\]
now find the common denominator
there isn't any common factor

Answer 10

LCD, so

\[\frac{ 1 }{ \cos^2 x \sin^2 x }\]

Answer 11

Actually numerator wouldn't be one.

Answer 12

right denominator is correct
remember when the denominators are different you should multiply
the top and bottom to make the same denominator

Answer 13

\[\rm \frac{ 1 }{ \cos^2x*sin^2x}-\frac{1}{\cos^2x}-\frac{1}{\sin^2x}+1\]
to get the same denominator for 2nd fraction what would you multiply by ?

Answer 14

sin^2x

Answer 15

correct multiply 1/cos^2x by sin^2/sin^2x \[\rm \frac{ 1 }{ \cos^2x*sin^2x}-\frac{1*\color{Red}{sin^2x}}{\cos^2x*\color{red}{sin^2x}}-\frac{1}{\sin^2x}+1\]

Answer 16

So:

\[\frac{ 1 }{ \cos^2 x \sin^2x } - \frac{ \sin^2x }{ \cos^2x \sin^2x } - \frac{ \cos^2x }{ \cos^2x \sin^2x } + \frac{ \cos^2x \sin^2x }{ \cos^2x \sin^2x }\]

I assume you can turn the one into a fraction to make it easier?

Answer 17

correct and yes now write it as a single fraction :=))

Answer 18

\[\frac{ 1 - \sin^2x - \cos^2x (\cos^2x) (\sin^2x) }{ \cos^2x \sin^2x }\]

Answer 19

Oops, plus.

Answer 20

\[\frac{ 1 - \sin^2x - \cos^2x +(\cos^2x) (\sin^2x) }{ \cos^2x \sin^2x }\]
like this got you

Answer 21

alright now use the identity sin^2x+cos^2x=1 solve for either sin or cos doesn't matter and then substitute that

Answer 22

Wait, is it just going to end up being 1?

Answer 23

maybe how did you get one ? ;)

Answer 24

Crossing out: the cos^2xsin^2x on the numerator and denominator cross out. Leaving:

\[1 - \sin^2x - \cos^2x\]

Answer 25

you're on a right track but no
you can't cancel out the sin^2 cos^2x first
bec 1-sin^2x-cos^2 is also divided by cos^2sin2

Answer 26

first solve this identity for sin or cos
sin^2x+cos^2x=1
so sin^2x= ??

Answer 27

1 - cos^2x

Answer 28

right we can replace sin^2x with 1-cos^2x \[\frac{ 1 - (1-sin^2x)- \cos^2x +(\cos^2x) (\sin^2x) }{ \cos^2x \sin^2x }\]
distribute by negative sign and then simplify

Answer 29

Ugh, sorry my internet went out. I'm back.

Answer 30

its okay you fine :=)

Answer 31

\[\frac{ -1 + \sin^2x - \cos^2x + (\cos^2x) (\sin^2x) }{ \cos^2x \sin^2x }\]

Answer 32

ohh wait idid a mistake it supposed to be 1-cos^2

Answer 33

\[\rm \frac{ 1 - (\color{Red}{1-cos^2x})- \cos^2x +(\cos^2x) (\sin^2x) }{ \cos^2x \sin^2x }\]
now distribute -(1-cos^2x)

Answer 34

it is same as -1(1-cos^2x)

Answer 35

-1 + cos^2x

Answer 36

right \[\rm \frac{ 1 - \color{Red}{1+cos^2x}- \cos^2x +(\cos^2x) (\sin^2x) }{ \cos^2x \sin^2x }\]
and now simplify(combine like terms )

Answer 37

\[\frac{ 1 + \cos^2x \sin^2x }{ \cos^2x \sin^2x }\]

Then they finally cross out?

Answer 38

well 1 is also cancels out

Answer 39

\[\rm \frac{ \cancel{1} \cancel{- \color{Red}{1}+\cancel{cos^2x}}- \cancel{\cos^2x} +(\cos^2x) (\sin^2x) }{ \cos^2x \sin^2x }\]

Answer 40

then u left with sin^2cos^2x/sin^2cos^2x

Answer 41

Okay! I get it now, thank you!

Answer 42

np yw :=)) good work!

Answer 43

well one of those lel go with np :P

Answer 44

Haha.