Question

Can someone check my answer, I feel like I did it completely wrong!!
13. Verify the identity. Justify each step.


sec (0) sec(0)
------------- - ------------- = 2csc * (0)
csc(0)-cot(0) csc(0)+cot(0)

(sec(0)*(csc(0)+ cot(0)) - (sec(0)*(csc(0)-cot(0)))
--------------------------------------------------- = 2csc*(0)
(csc(0)-cot(0))*(csc(0)*(csc(0)+cot(0))


(sec(0)*csc(0)+cot(0)*sec(0))-(2sec(0)*csc(0)-cot(0)*sec(0))
------------------------------------------------------------
(csc^2(0)-cot^2(0))
=2csc*(0)

(2csc*(2theta)+csc(0))-(2csc*(2theta)-csc(0))

Answer 1

sec (0) sec(0)
------------- - ------------- = 2csc * (0)
csc(0)-cot(0) csc(0)+cot(0)

Answer 2

(sec(0)*(csc(0)+ cot(0)) - (sec(0)*(csc(0)-cot(0)))
--------------------------------------------------- = 2csc*(0)
(csc(0)-cot(0))*(csc(0)*(csc(0)+cot(0))

Answer 3

(sec(0)*csc(0)+cot(0)*sec(0))-(2sec(0)*csc(0)-cot(0)*sec(0))
------------------------------------------------------------
(csc^2(0)-cot^2(0))
=2csc*(0)

Answer 4

(2csc*(2theta)+csc(0))-(2csc*(2theta)-csc(0))
--------------------------------------------- = 2csc*(0)
(csc^2(0)-cot^2(0))

Answer 5

csc^2(0)-cot^2(0)=1

((2csc*(2theta)+csc(0))-2csc*(2theta)+csc(0))

2csc(0)=2csc(0)

Answer 6

confusion and confusion o,o
how did you get 2 there http://prntscr.com/8qb8cf ?

Answer 7

im not quite sure. I watched a video about how to do it but i feel like i didnt learn a thing lol could you help me

Answer 8

hmm sure
i'll just start from http://prntscr.com/8qb8um
not gonna start over by different way

Answer 9

omg thank you! you are a lifesaver. my teacher already thinks im crazy with some of the answers i give her

Answer 10

distribute \[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\]
ignore theta for a sec

Answer 11

what's the definition of sec?
like csc = 1/sin
so sec = ?

Answer 12

im not sure:/

Answer 13

oh okay thats fine
but these are very important \[\sin \theta= \frac{1}{\csc \theta} ~~~\cos \theta =\frac{1}{\sec \theta }~~~ \tan \theta= \frac{1}{\cot \theta }\]
sin cos tan and csc sec cot are reciprocal

Answer 14

\[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\]
first look at the numerator
combine like terms !let me know what u get :=)

Answer 15

like sec*csc and -sec*csc are like terms
so sec*csc - sec*csc = ?

Answer 16

that is a different question right ?

Answer 17

i redid the equation

Answer 18

i'm confused don't know how u got tan and cos at the beginning

Answer 19

the work u posted here is almost correct but there was a little mistake so i'll say keep the same method

Answer 20

\[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\]
just two steps
then we are done !

Answer 21

oh shoot it was the wrong one im sorry. i guess ive been doing too much work today apparently lol ill just start from where you lfet off then

Answer 22

ohh i was so confused.

Answer 23

my baddd :)

Answer 24

its okay \[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\]
can you simplify the numerator just like simple algebra :=) combine like terms :=)

Answer 25

wouldnt it cancel eachother out because sec*csc+sec*cot is repeated twice
so it would be
sec*csc+sec*cot as the numerator

Answer 26

yes right sec*cot + sec *cot = 2 sec*cot
\[\huge\rm \frac{ 2\sec*\cot }{ \sec^2 - \cot^2}\]
we can convert sec to 1/cos and cot to cos/sin

Answer 27

cot = cos /sin
and tan = sin/cos
bot hare reciprocal of each other

Answer 28

oh well latex doesn't work