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rvc
 one year ago
Please help :)
rvc
 one year ago
Please help :)

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rvc
 one year ago
Best ResponseYou've already chosen the best response.1\[\rm Show~that~: \tan^{1}i(\frac{ xa }{ x+a })=\frac{ i }{ 2 }\log \frac{ x }{ a }\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0is \(x\) a complex number ?

rvc
 one year ago
Best ResponseYou've already chosen the best response.1i think so no it is not mentioned in the question

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2its not, i recently solved same/similar question...trying to recollect what i had done :P

rvc
 one year ago
Best ResponseYou've already chosen the best response.1do we have formula for \[2\tan^{1} x\]

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2one long way to do it is: write \(\tan^{1} iA = i\tanh^{1}A \) and use the formula for \(\tanh^{1}A\) \(\tanh^{1}p = (1/2) \ln (\dfrac{1+p}{1p})\)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2A(or p) = (xa)/(x+a) (1+p)/(1p) indeed gives you x/a

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2\(\tan^{1} iA = i\tanh^{1}A\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Have you studied of Hyperbolic Functions yet?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then, yes, you can do like what hartnn wrote above...

rvc
 one year ago
Best ResponseYou've already chosen the best response.1yay!!!!!!! solved :) @hartnn thanks

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0bit late, but alternative approach: let \(\large \theta = tan^{1} i (\frac{xa}{x+a})\) dw:1439128404910:dw \(\large e^{i\theta} = cis \ \theta \implies \theta = i \ ln \frac {1}{cis \ \theta}\) from \(\large tan \theta = \frac{i \ (xa)}{x+a}\) \(\large cos \theta = \frac{x+a}{\sqrt{4ax}}\) \(\large sin \theta = \frac{i \ (xa)}{\sqrt{4ax}}\) \(\large cis \ \theta = \frac{2a}{ \sqrt{4ax} } = \sqrt{\frac{a}{x}} \) \(\large \theta = i \ ln \sqrt{\frac{x}{a}} = \frac{i}{2} \ ln \frac{x}{a}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Very nice approach @IrishBoy123 ..
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