Question

how do i verify

(cos x/ 1 + sin x) + (1 + sin x/ cos x) = 2 sec x?

i don't know how to verify anything at all can someone explain how it works?

Answer 1

\[\huge\rm \frac{ \cos(x) }{ 1+\sin(x) } +\frac{ 1+\sin(x) }{ \cos(x) }\]
1)what is the common denominator ?

Answer 2

i dont know

Answer 3

i really dont know anything about this

Answer 4

alright let's change that to `simple` algebra

Answer 5

\[\huge\rm \frac{ x }{ y }+\frac{ y }{ x }\] what's the common denominator ?

Answer 6

xy

Answer 7

okay let's say y represent 1+sinx
and x = cos(x)
\[y=1+\sin(x)\]
\[x=\cos(x)\]
now can you tell me what would be the common denominator\[\huge\rm \frac{ \cos(x) }{ 1+\sin(x) } +\frac{ 1+\sin(x) }{ \cos(x) }\]
for this question :=)

Answer 8

1+sinx*cosx ?

Answer 9

PERFECT!

Answer 10

hi i speak Arabic

Answer 11

salam alakom

Answer 12

Nnesha

Answer 13

\[\huge\rm \frac{ \cos(x) }{ \color{red}{1+\sin(x)} } +\frac{ 1+\sin(x) }{\color{blue}{ \cos(x)} }\]
\[\huge\rm \frac{ \cos(\color{blue}{cos}) + (\color{reD}{1+\sin})(1+\sin)}{ \cos(x) (1+\sin(x))}\]

multiply the numerator of `1st fraction` by the `denominator` of 2nd fraction
multiply the numerator of `2nd fraction` with the denominator of `1st fraction`

Answer 14

so would it become cos^2x and (1+sin)^2 on the numerator??

Answer 15

yes right!!!

Answer 16

\[(1+\sin(x))^2 = (1+\sin)(1+\sin)\]
(1+sin)^2 is same as (1+sin)(1+sin) foil it

Answer 17

and what happens when i foil it?

Answer 18

you will get 3 terms :3

Answer 19

then you will be able to cancel out some stuff at the numerator :=)

Answer 20

so 1+2sinx+sin^2x ???

Answer 21

yes right \[\huge\rm \frac{ \color{ReD}{cos^2x}+1+2sinx+\sin^2x }{ cosx(1+sinx) }\]
cos^2x= what ?
familiar with the identity ?

Answer 22

\[\huge\rm sin^2 \theta+\cos^2\theta=1\] this one solve for cos^2

Answer 23

cos^2=1-sin^2 ?

Answer 24

replace cos^2 with 1-sin^2 and then simplify !

Answer 25

\[\huge\rm \frac{ \color{ReD}{1-sin^2x}+1+2sinx+\sin^2x }{ cosx(1+sinx) }\]

Answer 26

2+2sinx ?

Answer 27

yes!
what is common factor ?

Answer 28

\[\huge\rm \frac{ 2+2sinx }{ \cos(x)(1+sinx) }\]
2+sinx what is common in both terms?

Answer 29

sinx?

Answer 30

no
first term is 2 and 2nd one is 2sinx so sinx isn't common

Answer 31

so 2

Answer 32

yes right take out the 2 what will you get ?

Answer 33

sinx

Answer 34

nah

Answer 35

when you take out 2 from 2 what will you have left wth?

Answer 36

0

Answer 37

haha
well no
in other words you have to divide both terms with the common factor
so \[\huge\rm \frac{ 2}{ 2} +\frac{ 2sinx }{ 2 } =2(??+???)\]

Answer 38

1+sinx ?

Answer 39

yes!
and we should keep the common factor outside the parentheses \[\huge\rm \frac{ 2(1+sinx) }{ cosx(1+sinx) }\]

Answer 40

simplify! that

Answer 41

2/cosx

Answer 42

yes right and last step ?

Answer 43

i dont know

Answer 44

thats fine!
2/cosx is same as \[\huge\rm 2 \times \frac{ 1 }{ cosx }\]

Answer 45

and 1/cosx= ??

Answer 46

secx

Answer 47

so 2/cosx= ? :=)

Answer 48

EXCITING

Answer 49

i honestly have no idea how to do anything with identities so thank you

Answer 50

ikr! trig is FUN(well i hate graph)

Answer 51

my pleasure! :=)
welcome to openstudy!

Answer 52

\(\color{blue}{\text{Originally Posted by}}\) @rami2021
salam alakom
\(\color{blue}{\text{End of Quote}}\)
Walikum salam
and yeah arabic is my fvt language
but all i know is *kafya halik ?? ;~;

Answer 53

sorry about that alejand :3

Answer 54

no problem!

Answer 55

i'm about to post another question, i have one left