Question

prove that
ilog x-i/x+i= π-2tan-¹x

​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Concept:

Logarithms are another way of writing exponents in mathematics. A number's logarithm with a base equals another number. Exponentiation is the inverse function of a logarithm.

Given:

The given equation is $ilog[\frac{x-i}{x+i} ]=\pi -2tan^{-1} x$.

To prove:

$ilog[\frac{x-i}{x+i} ]=\pi -2tan^{-1} x$.

Solution:

Solving the left-hand side of the equation

∵ $log\frac{a}{b} = log(a)-log(b)$

$ilog[\frac{x-i}{x+i}]=i[log(x-i-log(x+i))]$

$=i[\frac{1}{2}log(x^{2} +1)-itan^{-1}\frac{1}{x} -\frac{1}{2}log(x^{2} +1)-itan^{-1}\frac{1}{x} ]$

(∵ $log(x+i)=\frac{1}{2}log(x^{2} +1)-itan^{-1}\frac{1}{x}$

and

$log(x-i)=\frac{1}{2}log(x^{2} +1)-itan^{-1}\frac{1}{x}$ )

Simplify further

$=i[-2itan^{-1}\frac{1}{x}]$

$=2tan^{-1}\frac{1}{x}$

$=2cot^{-1}x$

$=2[\frac{\pi}{2}-tan^{-1}x ]$

$=\pi -2tan^{-1}x$

which is equal to the right-hand side.

Hence proved.