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hartnnBest ResponseYou've already chosen the best response.1
use log. take log on both sides.
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
cos x * log cos x (cosx / cos x) * sinx + log cos x *  sinx
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
what is this ?? (cosx / cos x) * sinx + log cos x *  sinx
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
you can use LH only if you have 0/0 or infinity/infinity form
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Huh..i Just Forgot that..)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
this is same as x^x x>0 try this x^x = e^(x log(x))
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
so, log cos x/ sec x now try LH
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
because cos =1/sec that should work, i think....
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
that certainly works ... but this seems interesting ... using substitution. let cos(x) = u, then we have u > 0 , u^u ... i think you might have done before.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
something like sin /(sec cos tan) = cos log L = 0 L=1
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
you got 2 methods :) understood both ??
 one year ago
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