Question

Trig help. Attachment to follow.

Answer 1

My attempt at trying to simplify the first one doesn't seem to be going anywhere...i'm working on attaching a pic, just a sec.

Answer 2

sin square x plus cos square x is equal to 1

Answer 3

alrigh, but can i do anything with the rest of it?

Answer 4

\[\sin ^{2}x + \cos ^{2}x = 1\]

Answer 5

Use the identity :

\[\sin^2x + \cos^2x = 1\]

Answer 6

ok... looking at choice D

this is the expansion, you just need to finish it

\[\frac{ 1 - (\cos(x))^2}{(\cos(x)} = \frac{1 - 1 + \sin^2(x)}{\cos(x)} = \frac{\sin^2(x)}{\cos(x)} = \sin(x) \times \frac{\sin(x)}{\cos(x)}\]

Answer 7

@PoofyPenguin , the answer is
A-5
B-3
C-2
D-4
E-1

Answer 8

Could you please explain?

Answer 9

sure which part

Answer 10

don't forget that for A

\[\tan(x) = \frac{\sin(x)}{\cos(x)} = \sin(x) \times \frac{1}{\cos(x)}\]

whats 1/cos(x)

Answer 11

Alright, but the my solution that i posted initally, what do i do after the sin^2x+cos^2x=1 identity?

Answer 12

For the one u tried
use the identity
\(\cos^2x+\sin^2x=1\) as both @fall2012 and @waterineyes said.
then factorise the numerator.
Try to cancel common factors in both numerator and denominator if there are any.

Answer 13

Solving 1 part:

\[\frac{\cos^2x + 1 + \sin^2x - 2\sin^2x}{(1 - sinx)cosx} \implies \frac{2(1 - sinx)}{(1 - sinx)cosx} \implies 2\sec(x)\]

Answer 14

So E part gets combined with 1..

Answer 15

look at answer 3

take out the common factor

\[\sec(x)(1 - \sin^2(x)) = \sec(x)(\cos^2(x)) = \frac{1}{\cos(x)} \times \cos^2(x)\]

Answer 16

Part A is tan(x)
Also sin(x)sec(x)=sin(x)/cos(x)=tan(x)
hence A=5

Answer 17

thats enough from me... and hope it all helps..

Answer 18

Yes, it helped very much! Thank you! :D

Answer 19

Solving part 2 :

\[\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \implies \frac{\sin^2(x) + \cos^2(x)}{\sin(x) \cos(x)} \implies \frac{1}{\sin(x) \cos(x)} = Solve\]

Answer 20

And it is attaching to part C.

So : C is attaching to 2..

Answer 21

What else remains @PoofyPenguin

Answer 22

I'm still having issues with the one i initially posted... i can't seem to figure out how it was simplified

Answer 23

I have solved the same above in my second post...

Answer 24

Also remember
\[\frac{ 1 }{ sinx }=cscx\]
\[\frac{ 1 }{ cosx }=secx\]

Answer 25

I can't seem to figure out how you got to \[2\sin ^{2}x\]

Answer 26

\[\frac{\cos(x)}{1 - \sin(x)} + \frac{1 - \sin(x)}{\cos(x)} \implies \frac{\cos^2(x) + (1 - \sin(x))^2}{(1- \sin(x)) \cos(x)}\]
I have got this by taking LCM..
Getting ??

Answer 27

Oh sorry that is 2sinx not 2sin^2(x)

Answer 28

\[\frac{\cos^2(x) + \sin^2(x) + 1 - 2\sin(x)}{(1-\sin(x))\cos(x)} \implies \frac{2 - 2\sin(x)}{(1 - \sin(x))\cos(x)}\]

Answer 29

Yep! That's what i got so far, so i guess i would factor out the 2 next?

Answer 30

ya

Answer 31

Yep go for it...

Answer 32

Yay! I got it! Thanks again!

Answer 33

Welcome..

I said you no that ajprincess is here so don't worry..

Where I will be wrong she will there to correct me..

Ha ha ha..

Feeling very safe..

Answer 34

Ha ha ha!