Question

help

Answer 1

\[\frac{ sinx }{ 1-cotx}+\frac{ cosx }{ 1-tanx}=sinx+cosx\]

Answer 2

Really? :| oh man this one's even worse than the last lol.

Answer 3

I K

Answer 4

I think we'll have to do the same thing we did last time.. get a common denominator, apply a bunch of identities.. and then smash everything into tiny pieces until it looks like sine + cosine.

Answer 5

Errr nevermind, we don't want to combine them :) I think this will work out nicely if we keep them separate.

Answer 6

I dunno, let's work through it and find out :3

Answer 7

k then u could do some of the last stepscuse that is the part that i messed up

Answer 8

K AS U WISH!!

Answer 9

\[\large \left(\frac{1+\cot x}{1+\cot x}\right)\left(\frac{\sin x}{1-\cot x}\right)+\left(\frac{\cos x}{1-\tan x}\right)\left(\frac{1+\tan x}{1+\tan x}\right)\]

Answer 10

So we're multiplying the bottoms by their CONJUGATES, which allows for some nice simplification (hopefully...).

Answer 11

k

Answer 12

\[\large \frac{\sin x (1+\cot x)}{1-\cot^2x}+\frac{\cos x(1+\tan x}{1-\tan^2x}\]Hmm maybe this wasn't the right path to take actually... those bottoms aren't going to work out nicely like I had hoped :D one sec.. thinking,

Answer 13

k np

Answer 14

1st term by 1-tan ans second by 1-tan?

Answer 15

Mehhhh maybe...
Let's try this I guess..
We'll simplify 1-cot x a second, by converting to sines and cosines.\[\large 1-\cot x \quad = \quad \frac{\sin x}{\sin x}-\frac{\cos x}{\sin x}\quad \rightarrow \quad \frac{\sin x-\cos x}{\sin x}\]First term becomes,\[\huge \frac{\sin x}{1-\cot x}\quad \rightarrow \quad \frac{\sin x}{\frac{\sin x-\cos x}{\sin x}}\]\[\huge \sin x\cdot \frac{\sin x}{\sin x - \cos x} \quad \rightarrow \quad \frac{\sin^2x}{\sin x-\cos x}\]

Answer 16

Hmmm :3

Answer 17

Hmm this one might have me stumped :D

Answer 18

it is so hard

Answer 19

k just forget that stupid question help me with this one

Answer 20

\[\cos ^{4}x-\sin ^{4}x=\cos ^{2}x-\sin ^{2}x\]

Answer 21

I SWEAR THIS IS THE LAST ONE PLZ HELP

Answer 22

Oh this one might actually be fun :) heh

Answer 23

k :D

Answer 24

\[\large \cos^4x-\sin^4x \quad \text{we can write this as such,}\quad (\cos^2x)^2-(\sin^2x)^2\]

Answer 25

Notice that we have the difference of squares?

Answer 26

\[\large a^2-b^2=(a-b)(a+b)\]In our problem, a = cos^2x.

Understand how I broke up the 4th power like that?

Answer 27

k

Answer 28

We can write it as conjugates like so,\[\large (\cos^2x)^2-(\sin^2x)^2 \qquad \rightarrow \quad (\cos^2x-\sin^2x)(\cos^2x+\sin^2x)\]

Answer 29

\[\large (a)^2-(b)^2=(a-b)(a+b)\]We just applied this idea. Hopefully it's not confusing with all the powers floating around :)

Answer 30

k

Answer 31

Hmm this is interesting, we've been able to show so far that,\[\large (\cos^2x-\sin^2x)(\cos^2x+\sin^2x) \quad = \quad \cos^2x-\sin^2x\]
See how one of the sets of brackets is EXACTLY the thing on the right?
So from here, we just need to show that the other set of brackets is simply 1.

Answer 32

Is it jumping out at you yet? :D
It's probably the first identity you learn in trig! A very important one! heh

Answer 33

but how did u get cos^2 why no 4?

Answer 34

I broke up the powers on the trig functions like this.
\[\large x^4=(x^2)^2\]

Answer 35

yeah but in ur answer

Answer 36

There shouldn't be any more 4th powers after we broke them down into conjugates. Was that step kinda confusing? :o

Answer 37

I hope the trig exponent notation isn't confusing you. They're placed in a weird spot.\[\large \cos^4x=(\cos x)^4\]From there, we rewrote this as,\[\large (\cos x)^4 \quad \rightarrow \quad \left((\cos x)^2\right)^2\]

Answer 38

yes

Answer 39

is that it?

Answer 40

k thanks once again

Answer 41

is that it?