Question

trigonometry batado pls friends​

Answer 1

Answer:

$\cos(a) = \frac{1}{2} \\ then \: \cos(30) = \frac{1}{2} \\ a = 30 \\ \sin(b) = \frac{1}{ \sqrt{2} } \\ then \: \: b = 45 \\ we \: use \: the \: formula \\ tan(a - b) = \frac{ \tan(a) - \tan(b) }{1 + \tan(a) \tan(b) } \\ \frac{ \tan(30) - \tan(45) }{1 + \tan(30) \tan(45) } = \frac{ \frac{1}{ \sqrt{3} } - 1 }{1 + \frac{1}{ \sqrt{3} } \times 1 } \\ = \frac{ \frac{1 - \sqrt{3} }{ \sqrt{3} } }{ \frac{1 + \sqrt{3} }{ \sqrt{3} } } \\ = \frac{1 - \sqrt{3} }{1 + \sqrt{3} } \\ ratanaize \: it \\ \frac{1 - \sqrt{3} }{1 + \sqrt{3} } \times \frac{1 - \sqrt{3} }{1 - \sqrt{3} } \\ = \frac{1 + 3 - 2 \sqrt{3} }{1 - 3 } = \frac{2(2 - \sqrt{3)} }{ - 2 } \\ = - (2 - \sqrt{3} )$

it also solved by second method

$\tan(a - b) = \tan(30 - 45) \\ \tan( - 15 ) = - \tan(15) \\ \ - tan(15) = - (2 - \sqrt{3} )$

your answer