Question

prove that this problem

Answer 1

first of all we assume
$cos^{-1}\frac{1-a^2}{1-b^2}=A\\\\cos^{-1}\frac{1-b^2}{1+b^2}=B$
and then , find
$tan\frac{A}{2},tan\frac{B}{2}$
with help of formula,
$tan\frac{x}{2}=\sqrt{\frac{1-cosx}{1+cosx}}$
now, got
tanA/2 = a , and tanB/2 = b
now, use formula ,
$tan\frac{A-B}{2}=\frac{tanA/2-tanB/2}{1+tanA/2.tanB/2}$
after putting value of tanA/2 and TanB/2
we get,
$tan\frac{A-B}{2}=\frac{a-b}{1+ab}\\\\\frac{A-B}{2}=tan^{-1}\frac{a-b}{1+ab}\\\\A-B=2tan^{-1}\frac{a-b}{1+ab}$
now put A , and B value
then, you get
$cos^{-1}\frac{1-a^2}{1+a^2}-cos^{-1}\frac{1-b^2}{1+b^2}=2tan^{-1}\frac{a-b}{1+ab} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: hence \: proved$

Answer 2

ANSWER IS IN THE IMAGE