Question

prove the identity (show steps):
(tan(x)+1)^2=1+2sin(x)cos(x)/cos^2(x)

Answer 1

Left hand side
=[tan(x)+1]^2
=tan^2(x)+2tan(x)+1
=sin^2(x)/cos^2(x)+2sin(x)/cos(x)+1
=[sin^2(x)+2sin(x)cos(x)]/cos^2(x)+1
not=Right hand side.
so please check your question

Answer 2

wolfram alpha says its true, but wont explain why.

Answer 3

either you entered it incorrectly into wolfram, or you might want to recheck the equation....

as written, this is not an identity: (tan(x)+1)^2=1+2sin(x)cos(x)/cos^2(x)
let x=pi/4....
left side = 4
right side = 3

Answer 4

i checked it, it is exactly as written. still says true.

Answer 5

i copy/paste from your problem into wolfram... where does it say it's true?
http://www.wolframalpha.com/input/?i=%28tan%28x%29%2B1%29%5E2%3D1%2B2sin%28x%29cos%28x%29%2Fcos%5E2%28x%29

Answer 6

you missed some parentheses. (tan(x)+1)^2=(1+2sin(x)cos(x))/cos^2(x)
http://www.wolframalpha.com/input/?i=%28tan%28x%29%2B1%29^2%3D%281%2B2sin%28x%29cos%28x%29%29%2Fcos^2%28x%29

Answer 7

correction, I missed some, sorry.

Answer 8

hmmm... that's prolly why the first helper said this is not an identity also.... that's not what's in your original post....

Answer 9

my bad

Answer 10

ok.... is see now.... so that last one IS the identity in quesion, right?

(tan(x)+1)^2=(1+2sin(x)cos(x))/cos^2(x) ???

Answer 11

yes, except you cut off the last )

Answer 12

or not. this is confusing.

Answer 13

that is the right one

Answer 14

just to be clear, it's this: \(\large (tanx+1)^2=\frac{1+2sinxcosx}{cos^2x} \)

Answer 15

yes

Answer 16

looks like putting everything in terms of sine/cosine would be best here...
since the right side is basically in terms of sine/cosine, let's see if we can simplify that side first....

working on the right side, can you simplify it any further?

Answer 17

would making it (1+sin(2x)/cos^2(x) help?

Answer 18

i wouldn't go that direction.... when proving identities, i try to get everything in terms of sinx, cosx, etc. instead of sin(2x), cos(2x), etc...

this is what i'd do: \(\large \frac{1+2sinxcosx}{cos^2x}=\frac{1}{cos^2x}+\frac{2sinxcosx}{cos^2x}= \) ????

Answer 19

can u simplify that last step?

Answer 20

yes, but how did you get that from the original?

Answer 21

plain algebra: \(\large \frac{a+b}{c}=\frac{a}{c}+\frac{b}{c} \)

Answer 22

i mean from (tan(x)+1)^2?

Answer 23

we'll get to the left side soon....
we're working ONLY with the right side for now....

Answer 24

2tan(x)

Answer 25

1/cos^2(x)+2tan(x)

Answer 26

do you see how i got to this point:

\(\large \frac{1+2sinx{cosx}}{{cos^2x}}=\frac{1}{cos^2x}+\frac{2sinx\cancel{cosx}}{\cancel{cos^2x}^{cosx}}=sec^2x+2tanx \)

Answer 27

yes

Answer 28

ok.... good.... that's as far as we can get on the right side... agreed?

Answer 29

yes

Answer 30

now the left side is easy.....

Answer 31

expand \(\large (tanx+1)^2 \) = ?????

Answer 32

i see now, thanks

Answer 33

sure??? ok....

Answer 34

tan^2(x)+1+2tan(x)
tan^2(x)+1=sec^2(x)
2tan(x)+sec^2(x)

Answer 35

nicely done!!! :)

Answer 36

thanks for the help

Answer 37

yw... :)