Limit that involves the angle addition and subtraction of tangent

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wolly

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[math]\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}[/math]
I know that

[math]\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}[/math]
and

[math]\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}[/math]
Do You multiply each fraction with this (1-tan(a)tan(x))(1+tan(a)tan(x))?
I expanded the tangents and now I'm stuck.
 
wolly said:
[math]\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}[/math]
I know that

[math]\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}[/math]
and

[math]\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}[/math]
Do You multiply each fraction with this (1-tan(a)tan(x))(1+tan(a)tan(x))?
I expanded the tangents and now I'm stuck.
Click to expand...
The two answers you show are not equivalent (as seen by graphing them); so you'll need to show us your actual work in order for us to see where you went wrong. Everything you said so far sounds fine.
 
wolly said:
[math]\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}[/math]
I know that

[math]\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}[/math]
and

[math]\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}[/math]
Do You multiply each fraction with this (1-tan(a)tan(x))(1+tan(a)tan(x))?
I expanded the tangents and now I'm stuck.
Click to expand...
[imath]\displaystyle \lim_{x\rightarrow 0}\frac{\tan(a + x) + \tan(a - x) - 2\tan a}{x^2}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0}\frac{\tan(x + a) - \tan(x - a) - 2\tan a}{x^2}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0}\frac{\sec^2(x + a) - \sec^2(x - a)}{2x}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0}\frac{2[ \ \sec^2(x + a)\tan(x + a) - \sec^2(x - a)\tan(x - a) \ ]}{2}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0} \sec^2(x + a)\tan(x + a) - \sec^2(x - a)\tan(x - a) = \ ?[/imath]
 
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