Question

Simplify tan inverse of 2cosx-3sinx divided by 3cosx+2sinx

Answer 1

Answer:

$\tan {}^{ - 1} ( \frac{2 \cos(x) - 3 \sin(x) }{3 \cos(x) + 2 \sin(x) } ) \\ \\ = \tan {}^{ - 1} ( \frac{2 - 3 \tan(x) }{3 + 2 \tan(x) } ) \\ \\ = \tan {}^{ - 1} ( \frac{ \frac{2}{3} - \tan(x) }{1 + \frac{2}{3} . \tan(x) } ) \\ \\ let \: \: tan \: \alpha = \frac{2}{3} \\ \\ = \tan {}^{ - 1} ( \frac{ \tan( \alpha ) - \tan(x) }{1 + \tan( \alpha ) \tan(x) } ) \\ \\ = \tan {}^{ - 1} ( \tan( \alpha - x) ) \\ \\ = \alpha - x \\ \\ = \tan {}^{ - 1} ( \frac{2}{3} ) - x$