Question

Write the simplest form of ${\tan }^{-1}\left[\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right]$ if $\frac{a}{b}\tan x>-1$

Answer 1

${\tan }^{-1}\left[\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right]$

$={\tan }^{-1}\left[\frac{\frac{a\cos x-b\sin x}{b\cos x}}{\frac{b\cos x+a\sin x}{b\cos x}}\right]$

$={\tan }^{-1}\left[\frac{\frac{a\cos x}{b\cos x}-\frac{b\sin x}{b\cos x}}{\frac{b\cos x}{b\cos x}+\frac{a\sin x}{b\cos x}}\right]$

$={\tan }^{-1}\left[\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x}\right]$

$={\tan }^{-1}\frac{a}{b}-{\tan }^{-1}\left(\tan x\right)$

$={\tan }^{-1}\frac{a}{b}-x$