Results 1 to 9 of 9

Math Help - difficult identity

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    8

    difficult identity

    I've been working on this problem for over an hour, and I still am unable to verify it. I know it is true, because I plugged in a value for x and it works, but i don't know how to prove it.

    (tan x / (1 - cot x)) + (cot x / (1-tan x))^2 = 1 + sec x csc x

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qwerty723519@hotmail.com View Post
    I've been working on this problem for over an hour, and I still am unable to verify it. I know it is true, because I plugged in a value for x and it works, but i don't know how to prove it.

    (tan x / (1 - cot x)) + (cot x / (1-tan x))^2 = 1 + sec x csc x

    Thanks
    start with the left hand side. combine the fractions, change everything to sines and cosines and simplify as much as possible.

    then go to the left hand side, change everything to sines and cosines and show that you can simplify it to get the same thing
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2008
    Posts
    8
    yea, I know thats the method of solving them, and I was actually able to solve 19 out of the 20 I had to do, but this one just doesn't seem to simplify.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qwerty723519@hotmail.com View Post
    yea, I know thats the method of solving them, and I was actually able to solve 19 out of the 20 I had to do, but this one just doesn't seem to simplify.
    just to clarify, is the problem \frac {\tan x}{1 - \cot x} + \left( \frac {\cot x}{1 - tan x} \right)^2 = 1 + \sec x \csc x ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2008
    Posts
    8
    ⁡No it is actually:

    〖tan⁡x/(1-cot⁡x )〗+ cot⁡x/(1-tan⁡x )=1+ sec⁡〖x csc⁡x 〗

    ps. I typed the formula into microsoft word using the symbols, so that it was easier to visualize, but it didn't copy over the same. How did you write that formula?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qwerty723519@hotmail.com View Post
    ⁡No it is actually:

    〖tan⁡x/(1-cot⁡x )〗+ cot⁡x/(1-tan⁡x )=1+ sec⁡〖x csc⁡x 〗

    ps. I typed the formula into microsoft word using the symbols, so that it was easier to visualize, but it didn't copy over the same. How did you write that formula?
    hehe, what you write now is even more confusing than the last one. what do the ?'s mean?

    i used LaTeX. see the LaTeX tutorial here to see how
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2008
    Posts
    8
    whoops, sorry. It doesn't display ?s for me, but I'll try this LaTex thing, which might take me a little while.

    <br />
\frac {\tan x}{1 - \cot x} + \frac {\cot x}{1 - \tan x} = 1 + \sec x \csc x<br />

    whew, that was rough.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,811
    Thanks
    701
    Hello, qwerty!

    \frac {\tan x}{1 - \cot x} + \frac {\cot x}{1 - \tan x} \:= \:1 + \sec x \csc x

    We have: . \frac{\dfrac{\sin x}{\cos x}}{1 - \dfrac{\cos x}{\sin x}} + \frac{\dfrac{\cos x}{\sin x}}{1 - \dfrac{\sin x}{\cos x}}


    Multiply each fraction by \frac{\sin x\cos x}{\sin x\cos x}

    . . \frac{\sin x\cos x}{\sin x\cos x}\cdot\frac{\dfrac{\sin x}{\cos x}}{1 - \dfrac{\cos x}{\sin x}} \;+ \;\frac{\sin x\cos x}{\sin x\cos x}\cdot\frac{\dfrac{\cos x}{\sin x}}{1 - \dfrac{\sin x}{\cos x}} \;= . \frac{\sin^2\!x}{\sin x\cos x - \cos^2\!x} + \frac{\cos^2\!x}{\sin x\cos x - \sin^2\!x}

    . . = \;\frac{\sin^2\!x}{\cos x(\sin x-\cos x)} - \frac{\cos^2\!x}{\sin x(\sin x - \cos x)} \;=\;\frac{\sin^3\!x - \cos^3\!x}{\cos x\sin x(\sin x-\cos x)}


    Factor: . \frac{(\sin x-\cos x)(\sin^2\!x + \sin x\cos x + \cos^2\!x)}{\cos x\sin x(\sin x-\cos x)} \;=\;\frac{\sin x\cos x + \overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}}}{\cos x\sin x}

    . . = \;\frac{\sin x\cos x + 1}{\cos x\sin x} \;=\;\frac{\sin x\cos x}{\cos x\sin x} + \frac{1}{\cos x\sin x} \;=\;1 + \sec x\csc x

    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Mar 2008
    Posts
    8
    wow thank you so much! I would've spent another 2 hours on that to no avail, because I didn't even think of using difference of cubes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Difficult-ish Trig Identity
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 5th 2011, 08:54 PM
  2. Difficult Trig Identity Question
    Posted in the Trigonometry Forum
    Replies: 11
    Last Post: December 5th 2010, 07:11 AM
  3. difficult(for me) DE
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: April 3rd 2010, 05:49 AM
  4. Replies: 3
    Last Post: April 18th 2008, 06:20 PM
  5. difficult trig identity help plz!
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: November 3rd 2007, 11:21 AM

Search Tags


/mathhelpforum @mathhelpforum