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Question
If | x | < 1, then prove that
`2 tan^-1 "x" = tan^-1 ("2x"/(1 - "x"^2)) = sin^-1 ("2x"/(1 + "x"^2)) = cos^-1 ((1 - "x"^2)/(1 + "x"^2))`
Solution
Let tan-1x = y
Then, x = tan y
Now, `tan^-1 ("2x"/(1 - "x"^2)) = tan^-1 (("2 tan y")/(1 - tan^2 "y"))`
`= tan^-1 (tan 2"y")`
= 2y
= 2 tan-1x ......(1)
`sin^-1 ("2x"/(1 + "x"^2)) = sin^-1 (("2 tan y")/(1 + tan^2 "y"))`
`= sin^-1 (sin 2"y")`
= 2y
= 2 tan-1x ......(2)
`cos^-1 ((1 - "x"^2)/(1 + "x"^2)) = cos^-1 ((1 - tan^2 "y")/(1 + tan^2 "y"))`
`= cos^-1 (cos "2y")`
= 2y
`= 2 tan^-1 "x"` ......(3)
From (1), (2) and (3), we get
`2 tan^-1 "x" = tan^-1 ("2x"/(1 - "x"^2)) = sin^-1 ("2x"/(1 + "x"^2)) = cos^-1 ((1 - "x"^2)/(1 + "x"^2))`
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