Question

number 7 @AZ

Answer 1

i got
cosx-1/cosx
but then what

Answer 2

can u post the ss again?

Answer 3

remember the identity from the previous question? what does sec^2 (x) - 1 = ??

Answer 4

\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
i got
cosx-1/cosx
but then what
\(\color{#0cbb34}{\text{End of Quote}}\)

when you're telling us where you got to, it's kind of important to tell us what steps you took to get there
otherwise I kind of have to guess how you got there

Answer 5

sec^2 x = 1/cosx
1/sec^2x -1 = cosx -1
so
mutliplied together
cosx-1/cosx

Answer 6

uh no no no

come on, snow

you know the rules and that's cheating

you can't do

\(\dfrac{a}{b-c} \neq \dfrac{a}{b} - \dfrac{a}{c}\)

also you have ^2 and you're losing them

Answer 7

oh, oops lol

Answer 8

so just use the identity I told you about because it's the easiest way to go about it haha

we know that
\(\tan^2 (x) + 1 = \sec^2(x)\)

so what is \( 1 - \sec^2(x) = ?\)

and then we can replace it in the denominator and then simplify that fraction

Answer 9

-tan^2 x

Answer 10

are you sure?

just subtract 1 on both sides, what are you left with? :b

Answer 11

OH my bad

Answer 12

LOL

Answer 13

I meant to write, what is \(\sec^2(x) - 1\)

because that's what is in your question

Answer 14

ohh
tan^2 x

Answer 15

there you go

and now let's simplify \(\dfrac{\sec^2 (x)}{\tan^2(x)}\)

change both into sin and cos and simplify

Answer 16

sec^2 x is 1/cos^2 x
tan^2 x is sinx^2/cos^2 x

sinx^2 /cos^4 x
????

Answer 17

how can i simplify

Answer 18

snow snow flake flake

snow flake snow flake

snow flake flake snow

flake flake snow snow

what are you doing

Answer 19

????????????????

Answer 20

\(\sf \dfrac{\dfrac{snow}{flake}}{\dfrac{snow}{flake}} = \dfrac{snow}{flake} \div \dfrac{snow}{flake}\)


which is equal to

\(\sf\dfrac{snow}{flake} \times \dfrac{flake}{snow}\)

Answer 21

you can't forget how to divide fractions, snow

Answer 22

uh

Answer 23

OHHHHHHHHHH

Answer 24

oops lmaoooo

Answer 25

\( \dfrac{\sec^2 (x)}{\tan^2(x)} = \dfrac{\dfrac{1}{\cos^2(x}}{\dfrac{\sin^2(x)}{\cos^2(x)}}\)

Answer 26

hehe, done now
xdddddddddddddddd
<3
thank youuuuuuuuuuuu

Answer 27

you're welcome flakesnow!!!

Answer 28

higpnjpige
lmaooo