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Math Help - How to solve this identity?

  1. #1
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    How to solve this identity?

    tan(pi/4 + x) + tan(pi/4 - x) = 2sec2x

    I know that I must use compound angle formula for tan, but I'm still having heaps of trouble. Can someone show me the steps to solve this please?

    Thanks in advance!
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    tan(pi/4 + x) =cotx
    +
    tan(pi/4 - x)=-cotx

    =0

    hmm.. or maybe I am fool!
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  3. #3
    MHF Contributor Unknown008's Avatar
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    \tan(\frac{\pi}{4} + x) + \tan(\frac{\pi}{4} - x) = \dfrac{\tan(\frac{\pi}{4}) + \tan(x)}{1-\tan(\frac{\pi}{4})\tan(x)} + \dfrac{\tan(\frac{\pi}{4}) - \tan(x)}{1+\tan(\frac{\pi}{4})\tan(x)}

    Recall that \tan(\frac{\pi}{4}) = 1

    = \dfrac{1+ \tan(x)}{1-\tan(x)} + \dfrac{1 - \tan(x)}{1+\tan(x)}

    Combine the fractions.

    = \dfrac{(1+ \tan(x))^2 + (1-\tan(x))^2}{1-\tan^2(x)}

    Simplify;

    = \dfrac{(1+ 2\tan(x) + \tan^2(x)) + (1- 2\tan(x) + \tan^2(x))}{1-\tan^2(x)}

    = \dfrac{2+ 2\tan^2(x)}{1-\tan^2(x)}

    = \dfrac{2(1+ \tan^2(x))}{1-\tan^2(x)}

    = \dfrac{2\sec^2(x)}{1-\tan^2(x)}

    Hm.. time to convert into the basic ratios.

    = \dfrac{\frac{2}{\cos^2(x)}}{1-\frac{\sin^2(x)}{\cos^2(x)}}

    Multiply by cos/cos

    = \dfrac{\frac{2}{\cos^2(x)}}{1-\frac{\sin^2(x)}{\cos^2(x)}} \times \dfrac{\cos^2(x)}{\cos^2(x)}

    = \dfrac{2}{\cos^2(x)-\sin^2(x)}

    Now, recall that: \cos^2(x) - \sin^2(x) = \cos(2x)

    = \dfrac{2}{\cos(2x)}

    = 2\sec(2x)

    Shown!
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  4. #4
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    Quote Originally Posted by Also sprach Zarathustra View Post
    tan(pi/4 + x) =cotx
    +
    tan(pi/4 - x)=-cotx

    =0

    hmm.. or maybe I am fool!
    You are confusing it with \tan(\frac{\pi}{2} \pm x) = \mp\cot(x).
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  5. #5
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    Quote Originally Posted by doodofnood View Post
    I know that I must use compound angle formula for tan, but I'm still having heaps of trouble.
    If you want to avoid the compound angle identity for the tangent function, express everything in terms of
    sine and cosine and use the symmetric relation \cos\left(\frac{\pi}{4}-t\right) = \sin\left(\frac{\pi}{4}+t\right). It will be less involving.

    Spoiler:
    \displaystyle \tan\left(\frac{\pi}{4} + x\right) + \tan\left(\frac{\pi}{4} - x\right) = \frac{\sin\left(\frac{\pi}{4}+x\right)}{\cos\left(  \frac{\pi}{4}+x\right)}+\frac{\sin\left(\frac{\pi}  {4}-x\right)}{\cos\left(\frac{\pi}{4}-x\right)}
    Spoiler:
     \displaystyle = \frac{\sin^2\left(\frac{\pi}{4}+x\right)+\cos^2\le  ft(\frac{\pi}{4}+x\right)}{\cos^2\left(\frac{\pi}{  4}+x\right)} = \frac{1}{\frac{1}{2}\cos{2x}} = \frac{2}{\cos{2x}} = 2\sec{2x}.

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  6. #6
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    Hello, doodofnood!

    \tan(\frac{\pi}{4} + x) + \tan(\frac{\pi}{4} - x) \:=\: 2\sec2x

    \tan(\frac{\pi}{4} + x) + \tan(\frac{\pi}{4} - x) \;=\;<br />
\dfrac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4}\tan x} + \dfrac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4}\tan x}

    . . \displaystyle =\;\frac{1+\tan x}{1 - \tan x} + \frac{1-\tan x}{1+\tan x} \;=\;\frac{(1+\tan x)^2 + (1-\tan x)^2}{(1-\tan x)(1 + \tan x)}

    . . \displaystyle =\;\frac{1 + 2\tan x + \tan^2\!x + 1 - 2\tn x + \tan^2\!x}{1-\tan^2\!x} \;=\;\frac{2+2\tan^2\!x}{1-\tan^2\!x}

    . . \displaystyle =\;\frac{2(1+\tan^2\!x)}{1-\tan^2\!x} \;=\;\frac{2\sec^2\!x}{1-\tan^2\!x}


    \displaystyle \text{Multiply by }\frac{\cos^2\!x}{\cos^2\!x}\!:\;\;\frac{2}{\cos^2  \!x - \sin^2\!x} \;=\;\frac{2}{\cos2x} \;=\;2\sec2x



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