Question

Prove:

1/1+sinx+1/1-sinx=2sec^2x

1/1+sinx+1/1-sinx=...?

Answer 1

I'm confused.. what's your equation?

Answer 2

\[\dfrac{1}{1 + \sin(x)} + \dfrac{1}{1 - \sin(x)} = 2 \sec^2(x)\]

Answer 3

Try to get a common denominator on the left-hand side. Also try to pronounce the right-hand side loud. lol

Answer 4

\[\frac{ 1 }{ sinx+1 }+\frac{ 1 }{ 1-sinx }=2\sec^2x\]

Answer 5

I've done this proof so many times >.>

Answer 6

haha yes

Answer 7

\[\dfrac{1 - \sin(x) }{(1 + \sin(x))(1 - \sin(x))} + \dfrac{1 + \sin(x)}{(1 - \sin(x))(1 + \sin(x))}\]

Answer 8

You'd get something special in the denominator which is related to \(\sec^2(x)\).

Answer 9

and a 2

Answer 10

\[\dfrac{1 - \sin(x) + 1 + \sin(x)}{\text{I'd leave this part to you}}\]

Answer 11

SO THATS HOW YOU WRITE TEXT IN THERE omg.

Answer 12

@Luigi0210 Yeah, the 2 comes from the numerator.

Answer 13

Trig is so interesting :')

Answer 14

@isuckatmath9999

Answer 15

Trig loves you guys but hates me

Answer 16

im thinking -- err processing

Answer 17

Take your time.

Answer 18

my interesting version of math left a denominator of 1... lol

Answer 19

Naw

Answer 20

The denominator is \((1 - \sin(x))(1 + \sin(x))\). You can simplify that so easily.

Answer 21

normal people can simplify that easily lol

Answer 22

\[(x - y)(x + y) = x^2 - y^2 \]

Answer 23

Makes sense?

Answer 24

yes

Answer 25

So what would \((1 - \sin(x))(1 + \sin(x))\) be?

Answer 26

1^2-sin(x)^2
1-sin(x)^2
?

Answer 27

Indeed, indeed. See any identity there?

Answer 28

\[? = 1 - \sin^2(x)\]

Answer 29

uhhmm

Answer 30

Recall!

Answer 31

cosx
?

Answer 32

wawait that would need a sqrt

Answer 33

Yeah, close.

Answer 34

coverse sine?

Answer 35

Why are you complicating it?\[\sin^2(x) + \cos^2(x) = 1\]

Answer 36

xD

Answer 37

because I get easily confused?

Answer 38

Err man.

Answer 39

and i still hav eno idea why its 2sec^2x =/

Answer 40

\[\sin^2(x) + \cos^2(x) = 1 \Rightarrow \cos^2(x) = 1 - \sin^2(x)\]

Answer 41

You have\[\dfrac{1 + \sin(x) + 1 - \sin(x)}{1 - \sin^2(x)}\]Right?

Answer 42

right

Answer 43

Can you simplify the numerator?

Answer 44

I'm not sure how.
I don't have a chart of all of these things.

Answer 45

\[1 + \sin(x) + 1 - \sin(x)\]!!!

Answer 46

2

Answer 47

Good. And the denominator?

Answer 48

1-sin^2x?

Answer 49

Yeah, but I told you that \(1 - \sin^2(x) = \cos^2(x)\)

Answer 50

why do i need it to = cos^2x?

Answer 51

Because \(\sec^2(x)\) is related to \(\cos^2(x)\), RIGHT?

Answer 52

right?

Answer 53

\[\dfrac{2}{\cos^2(x)}\]is what you have

Answer 54

Now recall that\[\dfrac{1}{\cos^2(x)} = \sec^2(x)\]

Answer 55

ooohhhh

Answer 56

I get it!!! :D