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>> No.11562696 [DELETED]  [View]
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11562696

>>11562646
Typo, sorry. That whole line should be [math]
S-ba-ba^2=S\big[(b\tan\theta-ba\cos\theta)/(b\tan\theta)\big]=Sa^2 [/math]. There are no nested trig functions, so BE = b.tan t - b.sin t.cos t
>>11562619
Use trig to find CE. You should easily find it is b.sin t. Similarly, DE is b.(sin t)^2 and FE = b.(sin t)^3 and so on.
>>11562545
>>11562544
Pick [math] \epsilon=f'(a)/2 [/math]. Then by continuity of f' at a, we have for any delta [math] |x-a|<\delta\implies |f'(x)-f'(a)|<\epsilon [/math]. So there's [math] f'(x)-f'(a)<f'(a)/2 [/math] and [math] f'(a)-f'(x)<f'(a)/2 [/math]. Then [math] f'(x)>f'(a)/2=\epsilon [/math] which is a positive number.

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