Question

Please help :)

Answer 1

\[\rm Show~that~: \tan^{-1}i(\frac{ x-a }{ x+a })=\frac{ i }{ 2 }\log \frac{ x }{ a }\]

Answer 2

is \(x\) a complex number ?

Answer 3

i think so no
it is not mentioned in the question

Answer 4

its not, i recently solved same/similar question...trying to recollect what i had done :P

Answer 5

do we have formula for \[2\tan^{-1} x\]

Answer 6

one long way to do it is:
write \(\tan^{-1} iA = i\tanh^{-1}A \)
and use the formula for \(\tanh^{-1}A\)

\(\tanh^{-1}p = (1/2) \ln (\dfrac{1+p}{1-p})\)

Answer 7

@Michele_Laino

Answer 8

A(or p) = (x-a)/(x+a)

(1+p)/(1-p) indeed gives you x/a

Answer 9

what about i?

Answer 10

\(\tan^{-1} iA = i\tanh^{-1}A\)

Answer 11

Have you studied of Hyperbolic Functions yet?

Answer 12

yep

Answer 13

got you @hartnn

Answer 14

Then, yes, you can do like what hartnn wrote above...

Answer 15

yay!!!!!!!
solved :)
@hartnn thanks

Answer 16

welcome ^_^

Answer 17

bit late, but alternative approach:

let \(\large \theta = tan^{-1} i (\frac{x-a}{x+a})\)



\(\large e^{i\theta} = cis \ \theta \implies \theta = i \ ln \frac {1}{cis \ \theta}\)

from \(\large tan \theta = \frac{i \ (x-a)}{x+a}\)
\(\large cos \theta = \frac{x+a}{\sqrt{4ax}}\)
\(\large sin \theta = \frac{i \ (x-a)}{\sqrt{4ax}}\)
\(\large cis \ \theta = \frac{2a}{ \sqrt{4ax} } = \sqrt{\frac{a}{x}} \)
\(\large \theta = i \ ln \sqrt{\frac{x}{a}} = \frac{i}{2} \ ln \frac{x}{a}\)

Answer 18

thanks @IrishBoy123 :)

Answer 19

Very nice approach @IrishBoy123 ..