Question

Verify that this an identity:
(sec x - tan x)^2 + 1 all over sec x*csc x - tan x*csc x is equal to 2 tan x.

Answer 1

???

Answer 2

\[\huge \frac{(sec x - \tan x)^2 + 1}{\sec x \csc x - \tan x \csc x} \equiv 2\tan x\]
yes?

Answer 3

Yep! I keep getting really complicated fractions on the left side (obviously I would work on that side), can u guide me to 2tan x?

Answer 4

show me what you have

Answer 5

Um I wouldnt mind at all, nut it's kind of hard to type all of it on the iPod. My last line was (2 tan x sec x)/(sec x csc x - sec ). The denominator is what Is really screwing everything up, is there a way to simplify it?

Answer 6

*but, iPad

Answer 7

ahh i see it now. (i'll assume what you have done so far is right)

you have \[\huge \frac{2 \tan x \sec x}{\sec x \csc x - \sec x}\]
can you see that sec x is factorable in the denominator?

Answer 8

hmm now that i think about it... the numerator might be wrong...let me try solving it

Answer 9

Oh wait that's only half of it! That is the fraction being subtracted from (2 sec^2 x)/(sec x csc x - sec x)! Lol

Answer 10

ahh so you really have \[\Large \frac{2\sec^2 x}{\sec x \csc x - \sec x} - \frac{2\tan x}{\sec x \csc x - \sec x}\]
yes?

Answer 11

Yeah. I separated the fraction, but I don't think that's going to get me anywhere

Answer 12

* 2 tan x sec x

Answer 13

In the second frac

Answer 14

like i said...the sec x is factorable

\[\Large \implies \frac{2\sec^2 x}{\sec x(\csc x - 1)} - \frac{2 \tan x\sec x}{\sec x (\csc x - 1)}\]
yes?

Answer 15

Yes I saw that as well , but I erased it. Oh I see now that it can simplify the expression

Answer 16

indeed. so do you know the rest?

Answer 17

Well now it's down to 2sec x/ (csc x-1) - 2tan x/(csc x-1)
I would guess you would make tan x sin x over cos x, but what about the denominator?

Answer 18

good question

\[\csc x - 1\]
\[\implies \frac 1{\sin x} - \frac{\sin x}{\sin x}\]
\[\implies \frac{1 - \sin x}{\sin x}\]

Answer 19

Ok I'm down to 2sec x /(csc x -1) -2 sec x -2 tan x sin x, I'll get rid of denom

Answer 20

Thanks for the help, btw !You're really helpful :)

Answer 21

welcome

Answer 22

Ok im fraction free now, but what I have doesn't resemble 2tan x at all....
I have 2 sec x csc x - 4 sec x - 2 tan x sin x

Answer 23

\[\frac{2\sec x}{\csc x -1} - \frac{2\tan x}{\csc x - 1}\]
\[\implies \frac{\frac 2{\cos x}}{\frac{1-\sin x}{\sin x}} - \frac{\frac{2\sin x}{\cos x}}{\frac{1-\sin x}{\sin x}}\]
\[\frac{2\tan x}{1 - \sin x} - \frac{2\tan x \sin x}{1-\sin x}\]
is that what you did?

Answer 24

Yes! After that, you just bring the fractions together, factor out the 2 tan x, and the 1- sin x cancels out! Woohoo I got it! Thank you so much 4 ur help!

Answer 25

welcome