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

- anonymous

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

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- ParthKohli

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

- anonymous

i'll fix it

- anonymous

itit should be showing up properly now

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## More answers

- ParthKohli

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

- whpalmer4

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...

- ParthKohli

Ah, a very good idea.

- anonymous

you can't use both sides to prove it

- whpalmer4

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.

- ParthKohli

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.

- whpalmer4

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

- anonymous

ahh

- whpalmer4

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: \(\sin^2 x+ \cos^2 x = 1\)

- ParthKohli

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

- whpalmer4

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

- anonymous

^ I know that feeling

- ParthKohli

Me too :-P

- anonymous

Alright lemme see if I can get to the answer lol

- Jhannybean

Me 3.

- anonymous

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

- Jhannybean

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

- ParthKohli

LOL

- whpalmer4

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

- whpalmer4

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

- ParthKohli

iPad?
Richie rich alert.

- Jhannybean

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

- anonymous

man im so confuffled

- whpalmer4

Can you show us what you have?

- ParthKohli

@isuckatmath9999 Hint: 2 + 2 = 4.

- ParthKohli

Use that hint and you can get your answer.

- anonymous

How is that a hint?

- ParthKohli

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

- ParthKohli

Hehe. Just cross-multiply, eh?

- whpalmer4

Here, I'll show you the first two steps:
1) write "2+2"
2) write "=4"
:-)

- anonymous

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

- ParthKohli

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

- ParthKohli

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

- anonymous

>.>

- ParthKohli

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

- ParthKohli

The right-hand side, specifically.

- anonymous

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

- ParthKohli

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

- whpalmer4

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

- whpalmer4

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.

- anonymous

*twitch*

- whpalmer4

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

- anonymous

this isn't helping me at all. =/

- whpalmer4

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

- whpalmer4

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?

- whpalmer4

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

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