Question

Prove (tan 2x + cot 2x)/csc 2x = sec 2x
HELP!

Answer 1

I got to the step of
LHS = \[\frac{ (\frac{ \cos 2x }{ \sin 2x} + \frac{ \sin 2x }{ \cos 2x }) }{ \frac{ 1 }{ \sin 2x }}\]

Answer 2

okay.... and this is equal to sec(2x).

Answer 3

I'm confused at how to solve it after

Answer 4

when you are dividing by 1/sin(2x)

you are multiplying times sin(2x)

Answer 5

Yes it does, but I got to show all my work and simplify it.

Answer 6

\(\Large\color{black}{ \frac{\frac{\cos(2x)}{\sin(2x)}+\frac{\sin(2x)}{\cos(2x)}}{\frac{1}{\sin{2x}}}=\sec(2x) }\)


\(\Large\color{black}{ \sin(2x)(\frac{\cos(2x)}{\sin(2x)}+\frac{\sin(2x)}{\cos(2x)})=\sec(2x) }\)

Answer 7

right?

Answer 8

I got to keep going further all the way to 1/cos 2x

Answer 9

How do you multiply the sin 2x? put it on top of every fraction?

Answer 10

yes...

Answer 11

mutliply each top times sin(2x).

well... in a first fraction sin(2x) will cancel, and you will just get cos(2x).

Answer 12

then I get \[\cos 2x + \frac{ \sin^2 2x}{ \cos 2x }\]

Answer 13

yes.

Answer 14

now, re-write the cos(2x) as a fraction with a denominator of cos(2x).

Answer 15

set cos(2x)/1 and multiply top and bottom times cos(2x).

Answer 16

\(\Large\color{black}{ \frac{\cos(2x)}{1}+\frac{\sin^2(2x)}{\cos(2x)}=\sec(2x) }\)

Answer 17

multiply top and bottom of the first fraction times cos(2x), you get?

Answer 18

How does the left hand side end up to be \[\frac{ 1 }{ \cos 2x }\]

Answer 19

wait, you will see...

Answer 20

\(\Large\color{black}{ \frac{\cos(2x)}{1}+\frac{\sin^2(2x)}{\cos(2x)}=\sec(2x) }\)

\(\Large\color{black}{ \frac{\cos^2(2x)}{\cos(2x)}+\frac{\sin^2(2x)}{\cos(2x)}=\sec(2x) }\)

good with this?

Answer 21

recall that cos^2x+sin^2x=1

Answer 22

OH, I get it now, add the fractions and get 1/cos 2x = sec 2x!

Answer 23

yes... there you go:)

Answer 24

good work!

Answer 25

Want to help me with another to see if I get it?

Answer 26

yes... I can try:)

Answer 27

I'll make a new one

Answer 28

sure