Question

(____) ^2 = (secx -1) (secx +1)


Help with this?

Answer 1

@hartnn

Answer 2

there is a standard identity
\(\large \tan^2x=\sec^2x-1\)
write 1 as 1^2
on right side,now apply
\(a^2-b^2=(a+b)(a-b)\)
what u get ?

Answer 3

wait ...can you slow that down? 0.o

Answer 4

remember this standard identity
\(\large \tan^2x=\sec^2x-1\)

Answer 5

ok ?

Answer 6

Okay ..tan^2 x = sec^2 x -1 ..correct?

Answer 7

now you see sec^2 x -1 ?
write that as \(sec^2x-1^2\)
now it is of the form \(a^2-b^2\)
right ?

Answer 8

?

Answer 9

question tho.... it is that form but where is a2-b2 coming in at? 0.o

Answer 10

i thought , u would never return from chat.....

Answer 11

is that the secx + 1?

Answer 12

my bad. well im done in chat now

Answer 13

i seem to recall in algebra (x-2) (x+2) or somethign similar can be combined..right?

Answer 14

thats what the right side part made me think of ..but i ended up getting confused..hence my question on here..

Answer 15

\(sec^2x-1^2\)
is in the form of
\(a^2-b^2\)
and
\(a^2-b^2 = (a+b)(a-b)\)
so,
\(sec^2-1^2 = (sec x-1 )(secx+1)\)
got this ?

Answer 16

oh wait .... sec^2 -1^2 = (secx-1) (secx +1) right?

Answer 17

thats what i told.

Answer 18

yeah..its making sense now..but thne..thats the answer ? ...

Answer 19

and since , u know
\(\large \tan^2x=\sec^2x-1 \\ \large (\tan x)^2 = (secx-1)(sec x +1)\)

Answer 20

oh ..that would be the identity ....so the answer is tan x?

Answer 21

yes.

Answer 22

i get that one too

Answer 23

good :)

Answer 24

might i make a suggestion that people stop just giving out answers...it doesn't really help :/

Answer 25

they need to do it like you ...

Answer 26

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