The derivative of `tan^(-1)((sin x - cos x)/(sin x + cos x))`, with respect to ` x/2`, where `(x in(0,pi/2))` is

Options

A. 1
B. `2/3`
C. `1/2`
D. 2

Answer

D

Explanation

Let `f(x) = tan ^(-1) ((sin x - cos x)/(sin x + cos x)) = tan^(-1) ((tan x-1)/(tan x +1)) `
[dividing numerator and denominator by `cos x gt 0, x in (0, pi/2)`]
` = tan^(-1) ((tan x - tan pi/4)/(1+(tan pi/4) (tan x)))`
` = tan^(-1) [tan( x - pi/4)]`
`[:' (tan A-tanB)/(1+tan A tan B) = tan (A-B)]`
Since, it is given that ` x in (0, pi/2) `, so
` x - pi/4 in (-pi/4, pi/4)`
Also, for ` (x-pi/4) in (-pi/4, pi/4)`,
Than,
` f(x) = tan^(-1) (tan(x- pi/4)) = x - pi/4`
[`:' tan^(-1) tan theta=theta, " for " theta in (-pi/2, pi/2)`]
Now, derivative of ` f(x) w.r.t. x/2 ` is
`(d(f(x)))/(d(x//2))=2 (df(x))/(d(x))`
` = 2 xxd/(dx) (x-pi/4) =2`