Question

Write \(\tan^{-1}\left(\dfrac{\cos x-\sin x}{\cos x+\sin x}\right)\), \(\dfrac{-\pi}{4}<x<\dfrac{3\pi}{4}\) in the simplest form.

Hint:

Divide numerator and denominator by cos x to get (1−tanx)/(1+tanx)=tan(π/4−x).

Answer 1

Step 1: Key Formula or Approach:
Divide the numerator and denominator inside the fraction by \(\cos x\) to bring in \(\tan x\), then recognise the tangent subtraction formula \(\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}\) with \(A=\pi/4\) (since \(\tan(\pi/4)=1\)).

Step 2: Rewriting the fraction:
\[ \frac{\cos x-\sin x}{\cos x+\sin x} = \frac{1-\tan x}{1+\tan x} \]

Step 3: Matching to the tangent subtraction identity:
\[ \frac{1-\tan x}{1+\tan x} = \frac{\tan(\pi/4)-\tan x}{1+\tan(\pi/4)\tan x} = \tan\left(\frac{\pi}{4}-x\right) \]

Step 4: Applying \(\tan^{-1}\) and checking the domain:
\[ \tan^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right) = \frac{\pi}{4}-x \]
This step is valid because for \(-\pi/4<x<3\pi/4\), the angle \(\dfrac\pi4-x\) lies inside \((-\pi/2,\pi/2)\), the principal range of \(\tan^{-1}\), so no adjustment is needed.

Final Answer:
The simplest form is \(\dfrac{\pi}{4}-x\). \[ \boxed{\dfrac{\pi}{4}-x} \]