What kind of function is #f(x)=log((1+sinx)/(1-sinx))#? Whether odd, even or none?

1 Answer
Nov 1, 2017

#f(x)=log((1+sinx)/(1-sinx))# is an odd function.

Explanation:

A function is even if #f(-x)=f(x)#

and it is odd if #f(-x)=-f(x)#

here we have #f(x)=log((1+sinx)/(1-sinx))=log(1+sinx)-log(1-sinx)##

Hence #f(-x)=log((1+sin(-x))/(1-sin(-x)))#

= #log((1-sinx)/(1+sinx))# - (as #sin(-x)=-sinx#)

= #log(1-sinx)-log(1+sinx)#

= #-f(x)#

Hence #f(x)=log((1+sinx)/(1-sinx))# is an odd function.