Question

Precalc help please!
Prove 1 + sec^(2)x sin^(2)x = sec^(2)x

Answer 1

do you know this:
\(\sin^2x+\cos^2 x=1\)
?

Answer 2

yes that's the pythagorean identity

Answer 3

correct,
now try to isolate sin^2 x from that

(subtract cos^2 x from both sides)

what do u get ?

Answer 4

1 - cos^2x

Answer 5

correct,
now plug that in your left side equation
what u get?

Answer 6

1+sec^2 x (1-cos^2 x)

Answer 7

good,
distribute sec^2 x

\(1+\sec^2 x -\sec^2 x \cos^2 x\)
got this ?

Answer 8

can you simplify that further ?

Answer 9

how did we distribute sec2x?

Answer 10

\(\large ☻(☺+♥) = ☻☺+☻♥ \\ \text {this is distributing}\)

Answer 11

wait i got it!

Answer 12

\(\sec^2 x (1-cos^2 x ) = \sec^2 x - \sec^2 x \cos^2 x\)
good :)

Answer 13

can you tell me what sec^2 x cos^2 x = ... ?

Answer 14

this is where i get lost :(

Answer 15

did you know
\(\Large \sec x =\dfrac{1}{\cos x}\)

Answer 16

yes and cos x = 1/ sec x

Answer 17

which gives us
\(\Large \sec x \cos x = 1\)
isn't it ?

Answer 18

yes!

Answer 19

i think i got it!

Answer 20

1 + sec2x - 1
1 - 1 = 0
so, sec2x = sec2x

Answer 21

\(\huge \checkmark \)

Answer 22

AAHHHH! I can't believe I did it! Thank you soooooo much!

Answer 23

and since you are new here,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 24

you're most welcome ^_^

Answer 25

Can you check my work on one more problem? I feel like the process was too short to be right

Answer 26

sure

Answer 27

-tan^2 x + sec^2 x = 1
-tan^2 x + 1 + tan^2 x
the - tan and tan would cancel out so,
1 = 1

Answer 28

you wanted to prove
-tan^2 x + sec^2 x = 1
?

Answer 29

yes

Answer 30

ok, what you did is correct, but you have used sec^2 x = 1+ tan^2 x

Answer 31

if you are in doubts, don't use that

Answer 32

use these :
tan x = sin x/ cos x

sec x = 1/cos x

Answer 33

then try to simplify by taking the common denominator

Answer 34

so since tan is negative would it be - sinx/cosx or would i use the even odd identity

Answer 35

- tan x = -sin x/ cos x

-tan^2 x = - sin^2 x/ cos^2 x

Answer 36

alright that makes sense

Answer 37

so is this right so far,
- sin^2x/cos^2x + 1/cos^2x
1- sin^2x

Answer 38

denominator cos^2 x is common
so you combine the numerator

\(\Large \dfrac{1-\sin^2 x}{\cos^2x}\)

Answer 39

okay

Answer 40

and whats 1-sin^2 x = .. .?

Answer 41

cos^2x

Answer 42

good, so both numerator and denominator = cos^2 x
which gives you 1 :)

Answer 43

you're a genius!

Answer 44

not actually, but thanks for the compliment :)

Answer 45

anytime! :)