Question

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

Answer 1

We have,

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

On divide numerator and denominator by cosx and we get,

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

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{a\cos x-b\sin x}{\cos x}}{\frac{b\cos x+a\sin x}{\cos x}}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{a\cos x}{\cos x}-\frac{b\sin x}{\cos x}}{\frac{b\cos x}{\cos x}+\frac{a\sin x}{\cos x}}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{a-b\mathrm{tanx}}{b+a\tan x}\right)$

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

$\Rightarrow {\tan }^{-1}\left(\frac{\tan y-\tan x}{1+\tan y\tan x}\right)\therefore let\tan y=\frac{a}{b}$

$\Rightarrow {\tan }^{-1}\tan (y-x)\therefore So,y={\tan }^{-1}\frac{a}{b}$

$\Rightarrow (y-x)$

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

Hence, this is the answer.