Question

solve itttttt.......pzzzzzzzzzzzzzzzz


my dear frbda

Answer 1

#ANSWER #

$Solution:- \: \: \: \: \\ \\ Given \: \tan(A+B) = \sqrt{3} \\ = > \tan(A+B) = \sqrt{3} = \tan60° \\ = > \tan(A \: +B) = \tan60° \\ = > A+B \: = 60° \: \\ Also \: \: given \: \tan(A \: -B) = \frac{1}{ \sqrt{3} } = \tan30° \\ = > \tan(A-B ) = \tan30° \\ = > A-B = \: 30° \\ Adding \: (1) \: and \: (2) \: we \: get \: \\ 2A = 60°+ 30° = 90° \\ = > A = 90° \div 2 = 45° \\ \\ subtracting \: (2) \: from \: (1) \: we \: get \\ 2B = 60° - 30° = 30° \\ B = 30° \div 2 = 15° \\ = > Hence \: \: A = 45° \: And \: B = 15° \\ putting \: the \: values \: of \: A \: and \: B \: in \: \tan \: A. \sin(A+B) + \cos \:A. \tan(A - B) \: we \: get \\ \tan45° . \sin(45° + 15°) \cos45°. \tan(45° - 15°) \\ = tan \: 45°. \sin 60° + \cos \: 45°. \tan30° \\ = 1 \: . \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} } \: . \: \frac{1}{ \sqrt{3} } \\ = \frac{ \sqrt{3} }{2} \: . \frac{1}{ \sqrt{6} } = \frac{ \sqrt{18 \: } + 2}{2 \sqrt{6} } \\ = \frac{ 3\sqrt{2} + 2}{2 \sqrt{6} }$

THANX