Question

Need help with Trig Identities

(1-tan(x))/(1-cot(x))=-tan (x)

Answer 1

What is your question?

Answer 2

it didnt show up?

Answer 3

its (1-tan(x))/(1-cot(x)0=-tan(x)

Answer 4

sorry there shouldnt b a zero after cot (x) it was suppose to b a bracket

Answer 5

http://openstudy.com/study#/updates/501440bbe4b0fa246730bae2

Answer 6

So the fraction can be rewritten as: $$ \frac{1-tan(x)}{1-cot(x)} = \frac{1-\frac{sin(x)}{cos(x)}}{1- \frac{cos(x)}{sin(x)}} $$

Answer 7

Now multiply top and bottom by $$sin(x) \cdot cos(x)$$

Answer 8

$$\frac{1-\frac{sin(x)}{cos(x)}}{1- \frac{cos(x)}{sin(x)}} \cdot \frac{sin(x)cos(x)}{sin(x)cos(x)}$$

Answer 9

cant draw it to me cuz i cant see how ur doing that

Answer 10

y do u have to multiply it sin x * cos x

Answer 11

Because now the equation becomes:

$$ \frac{sin(x)cos(x) - sin^2(x)}{sin(x)cos(x) - cos^2(x)} $$

Answer 12

oh ok i see how u got there

Answer 13

now what do u do

Answer 14

The equation can be written as $$\frac{sin(x)(cos(x) - sin(x))}{cos(x)(sin(x) - cos(x))}$$

Answer 15

what happened to the squares? how did u get rid of them?

Answer 16

I pulled out a $$ sin(x) $$ from the top, and a $$cos(x)$$ from the bottom


Notice that if you distribute them, you will get back

$$\frac{sin(x)cos(x) - sin^2(x)}{sin(x)cos(x) - cos^2(x)}$$

Answer 17

w8 then shouldnt it be \[\frac{ 1*\cos(x)-\sin(x) }{ \sin(x)*1-\cos(x) }\]

Answer 18

if u take factor out sin (x) and cos (x)

Answer 19

Well yes, it's the same thing without the 1's

Answer 20

ok

Answer 21

You *could*, but then that would defeat the purpose of multiplying the fraction by $$sin(x)cos(x)$$ in the first place

Answer 22

So, now we can rewrite the fraction as :

$$\frac{sin(x)}{cos(x)} \cdot \frac{(cos(x) - sin(x))}{(sin(x) - cos(x))}$$

Answer 23

oh ok

Answer 24

And we know that $$\frac{sin(x)}{cos(x)}$$ is $$tan(x)$$

Answer 25

So the fraction is:

$$tan(x) \cdot \frac{cos(x) - sin(x)}{sin(x) - cos(x)}$$

Answer 26

oh ok then u can cancel out cos x-sin x/sin x-cos x which will give u -1

Answer 27

oh w8 no

Answer 28

Well, you can't cancel, because when you have a plus or minus, you can't cancel. You have to treat the top and bottom as one whole entity.

Answer 29

If you give me a minute, I can look up the identity required, but I know you can turn the second fraction into -1.

Answer 30

yea

Answer 31

Uggh. This is going to bug me. I know it's negative one, but I forget why.

Answer 32

i got it up to cosx(sinx)-1+sinx(cos x)

Answer 33

but i think that wrong

Answer 34

Oh... duh. Yeah.

Just like if you had:
$$\frac{(4-7)}{(7-4)} = \frac{-3}{3} = -1$$

This works the same way.

Answer 35

So the top and bottom will always be the same value, but will have different signs. Therefore, is will always be negative one.

Answer 36

it*

Answer 37

i dont got how to do get that with this

Answer 38

wat.

Answer 39

how do u get -1 with (cos x-sin x)/(sin x-cos x)

Answer 40

is there an identity for cosx(sinx)?

Answer 41

There is no trig identity or anything.

Say cos(x)-sin(x) = some value m
Then we can say sin(x)-cos(x) will be -m.

This is just like my example above:
if $$(7-4) = m$$
then
$$m=3$$

We also know that
$$(4-7) = -3$$

So you can see that when you flip the numbers in a subtraction problem, you will just get the negative result.
$$\frac{(4-7)}{(7-4)} = \frac{-m}{m} = \frac{-3}{3} = -1$$

Answer 42

Sorry if that's not clear enough. It's not anything complicated, I think you're overthinking it.

Answer 43

So:
If (a - b) = m.
We can conclude that (b - a) will be -m.

Answer 44

so u cant get -1 with out acutally solving the expression u just have to know its -1?

Answer 45

Yeah... one of the unfortunate parts of Calc & Trig.

Answer 46

well that sucks but ok thanks

Answer 47

Haha no problem. Hope it helped.

Answer 48

took 2 pages of writing for this question lol