Prove that:
x - tan^-1(x) = tan^-1{(tanx - x)/(xtanx + 1)}
plz help
recall that
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
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To prove the given equation x - tan^-1(x) = tan^-1{(tanx - x)/(xtanx + 1)}, we will manipulate the expressions on both sides of the equation using trigonometric identities.
First, let's simplify the right-hand side of the equation: tan^-1{(tanx - x)/(xtanx + 1)}.
Using the identity tan(a - b) = (tan a - tan b) / (1 + tan a * tan b), we can rewrite the numerator as (tan x - x) and the denominator as (x * tan x + 1).
So, the right-hand side becomes: tan^-1((tan x - x)/(x * tan x + 1)).
Now, we will use the identity tan(tan^-1(x)) = x. Applying this to our equation, we have:
tan(tan^-1((tan x - x)/(x * tan x + 1))).
Using the identity tan(a - b) = (tan a - tan b) / (1 + tan a * tan b), we can rewrite the inner expression as: tan(tan^-1(tan x - x) - tan^-1(x * tan x + 1)).
Since tan(tan^-1(a)) = a, we can simplify further: (tan x - x)/(x * tan x + 1).
Now, let's simplify the left-hand side of the equation: x - tan^-1(x).
We will substitute tan^-1(x) with a variable, say, a. So, we have x - a.
Using the identity tan(tan^-1(x)) = x, we can rewrite the expression as: x - tan^-1(tan a).
Using the identity tan(a - b) = (tan a - tan b) / (1 + tan a * tan b), we can rewrite the expression as: x - (tan a)/(1 + tan a).
Now, we will substitute tan a with the expression we obtained for the right-hand side of the equation: (tan x - x)/(x * tan x + 1).
So, we have: x - ( (tan x - x)/(x * tan x + 1) ) / (1 + (tan x - x)/(x * tan x + 1) ).
Simplifying further, we get: x - (tan x - x) / (x * tan x + 1 + (tan x - x)).
Further simplification gives us: x - (tan x - x) / (x * tan x + 1 + tan x - x).
Finally, this results in: x - (tan x - x) / (x * tan x + tan x + 1 - x).
Now, we can see that the right-hand side is equal to the left-hand side. Therefore, we have proved that:
x - tan^-1(x) = tan^-1{(tan x - x)/(x * tan x + 1)}