Question

here's your question dude

Answer 1

\[\frac{ \tan \beta-\sin \beta }{ \tan \beta+\sin \beta }=\frac{ \sec \beta-1 }{ \sec \beta+1 }\]

Answer 2

there you go prove that

Answer 3

I wasnt able to answer that

Answer 4

I like a challenge :)

Answer 5

You want me to simply prove it or guide YOU through proving it?

Answer 6

guide me...
It was a quiz of 3 questions I didnt answer any

Answer 7

Okay, give me a sec...

Answer 8

Bloody hell

Answer 9

Okay, start from the right-hand side
\[\Large \frac{\sec\beta -1}{\sec\beta+1}\]

Answer 10

And you can just multiply this by \[\large \frac{\sin \beta}{\sin\beta}\]
which is just 1...
Can you do that?

Answer 11

Hey, @EngrChong
hi...

Answer 12

Wait how was that?

Answer 13

How was what?

Answer 14

I thought you were only to solve for one side

Answer 15

Its either the lrft hand side or the right hand side

Answer 16

No, what you're supposed to do here is to take one side and somehow turn it into the other side by using identities or other such clever manipulations, like multiplying 1.

Answer 17

You could also start from the left-side and go in reverse... it's really up to you, but try the first method I posted...

Answer 18

What identities could you use?

Answer 19

You only need a pair of basic ones, actually...
\[\Large \sec\beta = \frac1{\cos\beta}\]and\[\Large \tan\beta = \frac{\sin\beta}{\cos\beta}\]

Answer 20

yup obviosly reciprocal and quotient identities

Answer 21

Yes, so let's take that right-hand side
\[\Large \frac{\sec \beta-1}{\sec\beta +1}\]
And multiply it by \(\Large \frac{\sin\beta}{\sin \beta}\)
And we get
\[\Large \frac{(\sin\beta)(\sec\beta -1)}{(\sin\beta)(\sec\beta +1)}\]
right?

Answer 22

yeah right

Answer 23

You get it? Okay, you distribute...
\[\Large \frac{(\sin\beta)(\sec\beta) - \sin\beta}{(\sin\beta)(\sec\beta)+ \sin\beta}\]
Can you already see where this is going?

Answer 24

Wait how did it become sin B?

Answer 25

These?
\[\Large \frac{(\sin\beta)(\sec\beta) - \color{red}{\sin\beta}}{(\sin\beta)(\sec\beta)+ \color{red}{\sin\beta}}\]

Answer 26

yup

Answer 27

Distribution ^_^
\[\Large \frac{(\sin\beta)(\sec\beta -1)}{(\sin\beta)(\sec\beta +1)}\]
remember that, say, in the numerator, sinB is distributed to both secB and the -1

Answer 28

oh yeah i forgot

Answer 29

Okay, so from here
\[\Large \frac{(\sin\beta)(\sec\beta) - \sin\beta}{(\sin\beta)(\sec\beta)+ \sin\beta}\]
It's pretty obvious where this is going, right?

Answer 30

Let me see

Answer 31

Cancel stuff?

Answer 32

\[\Large (\sin\beta)(\sec\beta) = (\sin\beta)\left(\frac{1}{\cos\beta}\right)\]\[\Large \color{white}{(\sin\beta)(\sec\beta)} = \left.\frac{\sin\beta}{\cos\beta}\right.\]\[\Large \color{white}{(\sin\beta)(\sec\beta)} = \tan\beta\]

Answer 33

Get it? :)

Answer 34

\[\frac{ \tan \beta-\sin \beta }{ \tan \beta+\sin \beta }\]

Answer 35

Yup. Precisely.

So, Wasn't that just fun? :)

Answer 36

Oh God it was that easy and wasnt able to answer it????????????The hell is wrong with me?

Answer 37

Nothing. It was easy, but it was FAR from obvious ^_^

Tips?

Answer 38

yeah sure...

Answer 39

Okay, when proving identities, you have your pencil and paper, you don't have to typeset stuff and all, so you can write complicated stuff pretty easy... so....

WHEN PROVING IDENTITIES, start with the more complicated side, say, the left side, since it has more trig functions... and THEN try to reduce everything ONLY in terms of sines and cosines.

Answer 40

There's actually two more questions

Answer 41

So... what you do here, let's try starting from the left...
\[\Large \frac{\tan\beta - \sin\beta}{\tan\beta +\sin\beta }\]

Answer 42

ok

Answer 43

It's not yet completely in terms of sines and cosines, you have tan. So let's replace tan with sin/cos
\[\Large \frac{\frac{\sin\beta}{\cos\beta}-\sin\beta}{\frac{\sin\beta}{\cos\beta}+\sin\beta}\]

Answer 44

I don't guarantee that expressing everything in sines and cosines gets you there right away, but it helps... it REALLY helps.

Here, you can see that you can actually factor out sinB :)
\[\Large \frac{(\sin\beta)\left(\frac1{\cos\beta}-1 \right)}{(\sin\beta)\left(\frac1{\cos\beta}+1 \right)}\]

Answer 45

We can cancel out the sinB, and we know that 1/cosB is equal to secB
\[\Large \frac{\cancel{(\sin\beta)}\left(\color{blue}{\sec\beta}-1 \right)}{\cancel{(\sin\beta)}\left(\color{blue}{\sec\beta }+1 \right)}\]

Answer 46

Voila
\[\Large \frac{\sec \beta-1}{\sec\beta +1}\]

Answer 47

how sin b got cancelled?

Answer 48

You can cancel any factors on the numerator AS LONG AS they have a partner in the denominator.

Here we had a factor of sinB in both numerator and denominator, so it cancels out.

Think numbers...
We know one third is the same as two sixths, and why? Well, because...
\[\Large \frac26 =\frac{2(1)}{2(3)}=\frac{\cancel2(1)}{\cancel2(3)}=\frac13 \]

Answer 49

oh i see

Answer 50

here's another one

Answer 51

\[\frac{ \sec \beta-\cos \beta }{ \sin^2 \beta }=\sec \beta \]

Answer 52

Easy >:)

Answer 53

But it requires a different method...

Answer 54

yup pythagorean

Answer 55

I'll admit that proving trig identities is tricky, because there is no ONE method to do it... you need experience... so let me demonstrate.

Answer 56

OH, so you already know how to answer this?

Answer 57

didnt complete it because of lack of confidence

Answer 58

Please show me what you did. I'll push you in the right direction if you stagger :)

Answer 59

it becme like this \[\frac{ \sec \beta-\cos \beta }{ 1-\cos^2 }\]

Answer 60

Okay... we could work from that... do you have a next step?

Answer 61

none got stuck there

Answer 62

I see. Could you turn the numerator into 1-cos^2B ?

Answer 63

Just remember...
\[\Large (\cos\beta)(\sec\beta) = 1\]

Answer 64

The tricky step here is to multiply the left-side by \(\Large \frac{\cos\beta}{\cos\beta}\)
^_^