rvc
  • rvc
Please help :)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rvc
  • rvc
\[\rm Show~that~: \tan^{-1}i(\frac{ x-a }{ x+a })=\frac{ i }{ 2 }\log \frac{ x }{ a }\]
ganeshie8
  • ganeshie8
is \(x\) a complex number ?
rvc
  • rvc
i think so no it is not mentioned in the question

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

hartnn
  • hartnn
its not, i recently solved same/similar question...trying to recollect what i had done :P
rvc
  • rvc
do we have formula for \[2\tan^{-1} x\]
hartnn
  • hartnn
one long way to do it is: write \(\tan^{-1} iA = i\tanh^{-1}A \) and use the formula for \(\tanh^{-1}A\) \(\tanh^{-1}p = (1/2) \ln (\dfrac{1+p}{1-p})\)
rvc
  • rvc
@Michele_Laino
hartnn
  • hartnn
A(or p) = (x-a)/(x+a) (1+p)/(1-p) indeed gives you x/a
rvc
  • rvc
what about i?
hartnn
  • hartnn
\(\tan^{-1} iA = i\tanh^{-1}A\)
anonymous
  • anonymous
Have you studied of Hyperbolic Functions yet?
rvc
  • rvc
yep
rvc
  • rvc
got you @hartnn
anonymous
  • anonymous
Then, yes, you can do like what hartnn wrote above...
rvc
  • rvc
yay!!!!!!! solved :) @hartnn thanks
hartnn
  • hartnn
welcome ^_^
IrishBoy123
  • IrishBoy123
bit late, but alternative approach: let \(\large \theta = tan^{-1} i (\frac{x-a}{x+a})\) |dw:1439128404910:dw| \(\large e^{i\theta} = cis \ \theta \implies \theta = i \ ln \frac {1}{cis \ \theta}\) from \(\large tan \theta = \frac{i \ (x-a)}{x+a}\) \(\large cos \theta = \frac{x+a}{\sqrt{4ax}}\) \(\large sin \theta = \frac{i \ (x-a)}{\sqrt{4ax}}\) \(\large cis \ \theta = \frac{2a}{ \sqrt{4ax} } = \sqrt{\frac{a}{x}} \) \(\large \theta = i \ ln \sqrt{\frac{x}{a}} = \frac{i}{2} \ ln \frac{x}{a}\)
rvc
  • rvc
thanks @IrishBoy123 :)
anonymous
  • anonymous
Very nice approach @IrishBoy123 ..

Looking for something else?

Not the answer you are looking for? Search for more explanations.