Question:
Write the function in the simplest form:
$\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right), 0
Solution:
$\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
$=\tan ^{-1}\left(\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}\right)$
$=\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$
$=\tan ^{-1}(1)-\tan ^{-1}(\tan x) \quad\left(\tan ^{-1} \frac{x-y}{1+x y}=\tan ^{-1} x-\tan ^{-1} y\right)$
$=\frac{\pi}{4}-x$