Question

Simplify
$tan^{-1}[ \frac{a\ cosx-b\ sinx}{b\ cosx+a\ sinx}]$ if a/b tan x > -1.

Answer 1

Please see the attachments

Answer 2

ANSWER:-

WE HAVE,

$\longrightarrow$$\huge\sf\red{{tan}^{ - 1}( \frac{a \: cos \: x - b \: sin \: x}{b \: cos \: x \: + a \: sin \: x} )}$

$\longrightarrow$$\huge\sf\red{ {tan}^{ - 1} ( \frac{ \frac{a \: cos \: x - b \: sin \: x}{b \: cos \: x} }{ \frac{b \: cos \: x + a \: sin \: x}{b \: cos \: x} } )}$

$\longrightarrow$$\huge\sf\red{ {tan}^{ - 1} ( \frac{ \frac{a}{b} - tan \: x }{1 + \frac{a}{b}tan \: x } )}$

$\longrightarrow$$\large\sf\red{ {tan}^{ - 1} \frac{a}{b} - {tan}^{ - 1} (tan \: x) = {tan}^{ - 1} \frac{a}{b} - x}$