anonymous
  • anonymous
Prove: tanx/(secx+1)=(secx-1)/tanx tanx/secx+1=?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ParthKohli
  • ParthKohli
Is this the equation?\[\dfrac{\tan(x)}{\sec(x) }+1 = \dfrac{\sec(x) - 1}{\tan(x)}\]
anonymous
  • anonymous
i'll fix it
anonymous
  • anonymous
itit should be showing up properly now

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More answers

ParthKohli
  • ParthKohli
Though you may keep in mind that \(\tan(x) = \dfrac{\sin(x)}{\cos(x)}\) and \(\sec(x) = \dfrac{1}{\cos(x)}\)
whpalmer4
  • 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
  • ParthKohli
Ah, a very good idea.
anonymous
  • anonymous
you can't use both sides to prove it
whpalmer4
  • 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
  • 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
  • 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
  • anonymous
ahh
whpalmer4
  • 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
  • ParthKohli
You can even use the identity \(\tan^2(x) = 1 - \sec^2(x)\) directly.
whpalmer4
  • whpalmer4
I'd have to prove that one to myself before using :-) Never did like memorizing trig identities!
anonymous
  • anonymous
^ I know that feeling
ParthKohli
  • ParthKohli
Me too :-P
anonymous
  • anonymous
Alright lemme see if I can get to the answer lol
Jhannybean
  • Jhannybean
Me 3.
anonymous
  • anonymous
so i somehow got to sin(x)/sin^2x+cos^2x
Jhannybean
  • 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
  • ParthKohli
LOL
whpalmer4
  • whpalmer4
well, you need a bit more than that, there's no = sign!
whpalmer4
  • whpalmer4
The URL for a trig identity page on my iPad takes up much less space than my old calc book :-)
ParthKohli
  • ParthKohli
iPad? Richie rich alert.
Jhannybean
  • Jhannybean
I can use my calc book as a weapon. -_- it's gigantic.
anonymous
  • anonymous
man im so confuffled
whpalmer4
  • whpalmer4
Can you show us what you have?
ParthKohli
  • ParthKohli
@isuckatmath9999 Hint: 2 + 2 = 4.
ParthKohli
  • ParthKohli
Use that hint and you can get your answer.
anonymous
  • anonymous
How is that a hint?
ParthKohli
  • 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
  • ParthKohli
Hehe. Just cross-multiply, eh?
whpalmer4
  • whpalmer4
Here, I'll show you the first two steps: 1) write "2+2" 2) write "=4" :-)
anonymous
  • 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
  • ParthKohli
\[\tan^2(x) = (\sec(x) - 1)(\sec(x) + 1)\]Simplify the right hand side using a well-known identity
ParthKohli
  • ParthKohli
You shouldn't write tan(x) = sin(x)/cos(x) right from the first step. That'd make things harder for you.
anonymous
  • anonymous
>.>
ParthKohli
  • ParthKohli
As I said, simplify \(\tan^2(x) = (\sec(x) - 1)(\sec(x) + 1)\)
ParthKohli
  • ParthKohli
The right-hand side, specifically.
anonymous
  • anonymous
theres no tan^2x on the right side is there? what?
ParthKohli
  • ParthKohli
Uh no. I'm telling you to simplify \((\sec(x) - 1)(\sec(x) + 1)\)
whpalmer4
  • 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
  • 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
  • anonymous
*twitch*
whpalmer4
  • whpalmer4
Or don't. If you don't, how about multiplying everything by \(\cos^2 x\) and see what happens?
anonymous
  • anonymous
this isn't helping me at all. =/
whpalmer4
  • whpalmer4
Do you see how I got to \[\frac{\sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} - 1\]?
whpalmer4
  • 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
  • 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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