Question

DOES ANYONE KNOW HOW TO SIMPLIFY THESE>????
#1)tan x(cos x – csc x)
#2)(sin x + cos x)(sec x+csc x)

Answer 1

great

Answer 2

first you gotta know some of the identities....

Answer 3

\[\tan x = { \sin x \over \cos x}\]

Answer 4

\[\sec x = { 1 \over \cos x}\]

Answer 5

\[\csc x = { 1 \over \sin x}\]

Answer 6

Lets do number one

Answer 7

alright

Answer 8

change the tan and the csc to the things i just taught you

Answer 9

I get:
\[{\sin x \over \cos x} (\cos x - {1 \over \sin x})\]

Answer 10

then distribute in

Answer 11

thats what i got

Answer 12

\[\sin x - {1 \over \sin x}\]

Answer 13

right

Answer 14

I prefer that form... your teacher may like csc instead

Answer 15

either way you are finished...

Answer 16

It's - 1/cos = sec

Answer 17

http://en.wikipedia.org/wiki/Trigonometric_functions#Reciprocal_functions

Answer 18

what about the second one??

Answer 19

What I posted was correct

Answer 20

ok thats what i thought

Answer 21

For the second one lets leave in sec x and cscx for now and treat it just like algebra

Answer 22

Sorry to interrupt. You made a mistake multiplying sin/cos times 1/sin in the first problem. It would be 1/cos

Answer 23

ahh yes you are correct! sorry...

\[\sin x - {1 \over \cos x}\]

Answer 24

I'll leave you to it. You're doing a great job.

Answer 25

@eyust707 small correction to #1 the solution is \[\sin x - \sec x\]
just recheck

Answer 26

ummmm ok now everyone is getting me confused :(

Answer 27

=P I made a little mistake on this one:

\[{\sin x \over \cos x} (\cos x - {1 \over \sin x})\]

Answer 28

Notice when you distribute the \(\sin x \over \cos x \) to both terms you end up with

\[\sin x - {1 \over \cos x}\]

not the other expression i posted previously

Answer 29

yes

Answer 30

okay so thats #1

Answer 31

#2 you just follow the rules of algebra

Answer 32

alright good :)

Answer 33

Pretend A B C D are the trig functions of that problem