Simplify tan ^{-1}left[frac{a cos x-b sin x}{ | vedclass

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(C) Given expression: $	an ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}ight]$Divide the numerator and denominator by $b \cos x$:$= 	an ^

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$	an ^{-1} \frac{a}{b}-x$

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(C) Given expression: $	an ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}ight]$Divide the numerator and denominator by $b \cos x$:$= 	an ^{-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}}ight]$$= 	an ^{-1}\left[\frac{\frac{a}{b} - 	an x}{1 + \frac{a}{b} 	an x}ight]$Using the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1}\left(\frac{A-B}{1+AB}ight)$,where $A = \frac{a}{b}$ and $B = 	an x$:$= 	an ^{-1}\left(\frac{a}{b}ight) - 	an ^{-1}(	an x)$$= 	an ^{-1}\left(\frac{a}{b}ight) - x$

Simplify $\tan ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right],$ if $\frac{a}{b} \tan x > -1$.

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