Question

Prove: tanx/(secx+1)=(secx-1)/tanx

tanx/secx+1=?

Answer 1

Is this the equation?\[\dfrac{\tan(x)}{\sec(x) }+1 = \dfrac{\sec(x) - 1}{\tan(x)}\]

Answer 2

i'll fix it

Answer 3

itit should be showing up properly now

Answer 4

Though you may keep in mind that \(\tan(x) = \dfrac{\sin(x)}{\cos(x)}\) and \(\sec(x) = \dfrac{1}{\cos(x)}\)

Answer 5

I think it is \[\frac{\tan x}{\sec x + 1} = \frac{\sec x - 1}{\tan x}\]

I would cross multiply to get the ball rolling...

Answer 6

Ah, a very good idea.

Answer 7

you can't use both sides to prove it

Answer 8

Sure you can. You manipulate the equation in any legitimate fashion (in other words, do the same thing to both sides) until you end up with an equation which is clearly true.

Answer 9

You can. Assume that it is true, and then use them. If you get to a point where everything is obvious, then it is true.

Answer 10

Similarly, if you end up at a point where everything is obviously not true, then you've proven that it is not true.

Answer 11

ahh

Answer 12

My rough sketch of how to prove this one, without remembering any obscure identities:

cross-multiply (notice you get a difference of two squares for one side)
rewrite sec as 1/cos
rewrite tan as sin/cos
multiply through by what hopefully is an obvious choice
use the most basic trig identity: \(\sin^2 x+ \cos^2 x = 1\)

Answer 13

You can even use the identity \(\tan^2(x) = 1 - \sec^2(x)\) directly.

Answer 14

I'd have to prove that one to myself before using :-) Never did like memorizing trig identities!

Answer 15

^ I know that feeling

Answer 16

Me too :-P

Answer 17

Alright lemme see if I can get to the answer lol

Answer 18

Me 3.

Answer 19

so i somehow got to sin(x)/sin^2x+cos^2x

Answer 20

I keep my calculus book close to my heart...only because it's got all the trig identities
and stuff in back, lolol.

Answer 21

LOL

Answer 22

well, you need a bit more than that, there's no = sign!

Answer 23

The URL for a trig identity page on my iPad takes up much less space than my old calc book :-)

Answer 24

iPad?

Richie rich alert.

Answer 25

I can use my calc book as a weapon. -_- it's gigantic.

Answer 26

man im so confuffled

Answer 27

Can you show us what you have?

Answer 28

@isuckatmath9999 Hint: 2 + 2 = 4.

Answer 29

Use that hint and you can get your answer.

Answer 30

How is that a hint?

Answer 31

You can use 2 + 2 = 4, then derive arithmetic.
Then derive algebra from it.
Then derive trigonometry from it, and solve your question :3

Answer 32

Hehe. Just cross-multiply, eh?

Answer 33

Here, I'll show you the first two steps:

1) write "2+2"
2) write "=4"

:-)

Answer 34

I have some gibberish of this
tan(x)/(secx-1)
=(sin(x)/cos(x))*(cos(x)/1)
=(sin(x)/1)
=sin(x)/sin^2x+cos^2x

Answer 35

\[\tan^2(x) = (\sec(x) - 1)(\sec(x) + 1)\]Simplify the right hand side using a well-known identity

Answer 36

You shouldn't write tan(x) = sin(x)/cos(x) right from the first step. That'd make things harder for you.

Answer 37

>.>

Answer 38

As I said, simplify \(\tan^2(x) = (\sec(x) - 1)(\sec(x) + 1)\)

Answer 39

The right-hand side, specifically.

Answer 40

theres no tan^2x on the right side is there? what?

Answer 41

Uh no. I'm telling you to simplify \((\sec(x) - 1)(\sec(x) + 1)\)

Answer 42

Seriously, after cross multiplying, you should have
\[\tan^2 x = (\sec x -1)(\sec x + 1) = \sec^2 x - 1\]\[\tan^2 x = \frac{1}{\cos^2 x} - 1\]Now plug in \[\tan x = \frac{\sin x}{\cos x}\]\[\frac{\sin^2x}{\cos^2x} =\frac{1}{\cos^2 x} - 1 \]and it should be easy from there

Answer 43

Just pretend you're simplifying \[\frac{s^2}{c^2} = \frac{1}{c^2}-1\] and replace s and c by sin and cos when you're done.

Answer 44

*twitch*

Answer 45

Or don't. If you don't, how about multiplying everything by \(\cos^2 x\) and see what happens?

Answer 46

this isn't helping me at all. =/

Answer 47

Do you see how I got to \[\frac{\sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} - 1\]?

Answer 48

I'll start again at the top, and you tell me after each step if you've got it:

\[\frac{\tan x}{\sec x + 1} = \frac{\sec x - 1}{\tan x}\]We cross-multiply:
\[\tan x * \tan x = (\sec x -1)(\sec x + 1) \]\[\tan^2 x = \sec^2x-\sec x + \sec x -1 = \sec^2 x - 1\]\[\tan^2 x = \sec^2 x -1\]

Okay so far?

Answer 49

Substitute \[\sec x = \frac{1}{\cos x}\]\[\tan^2 x = \frac{1}{\cos^2 x} - 1\]
Substitute \[\tan x = \frac{\sin x}{\cos x}\]\[\frac{\sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} -1\]Multiply everything by \(\cos^2x\)
\[\frac{\sin^2x}{\cancel{\cos^2x}}*\cancel{\cos^2x} = \frac{1}{\cancel{\cos^2x}}*\cancel{\cos^2 x} - \cos^2 x\]\[\sin^2 x= 1 -\cos^2x\]Add \(\cos^2x\) to both sides
\[\sin^2x + \cos^2 x = 1\checkmark\]