Question

Simplify...


Need explaination :)



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Answer 1

SOLUTION......

$\tan \: inverse \: ( \frac{a \cos(x) - b \sin(x) }{b \cos(x) + a \sin(x) } )$

Now ,

Dividing numerator and denominator by b cos x .

$\tan \: inverse \: ( \frac{ \frac{a \cos(x) - b \sin(x) }{b \cos(x) } }{ \frac{b \cos(x) + a \sin(x) }{b \cos(x) } } )$

$\tan \: inverse \: ( \frac{ \frac{a}{b} - \tan(x) }{1 + \frac{a}{b} \tan(x) } )$

We Know that :-

$tan \: a \: - tan \: b = \frac{tan \: a \: - tan \: b}{1 + tan \: a \: \times tan \: b}$

So,

$\tan \: inverse \: ( \frac{a}{b}) - \tan \: inverse( \tan(x) ) \:$

$\tan \: inverse \: ( \frac{a}{b} ) - x$

Answer 2

Answer:

$\huge\pink{hey\:mate\:refer\:to\:attatchment}$