Answer:
-33/56
Step-by-step explanation:
suppose: A,B are the 2 angles of a triangle
we have: A = arcsin[tex]\frac{-4}{5}[/tex]
B = arctan [tex]\frac{5}{12}[/tex]
=> sin A = [tex]\frac{-4}{5}[/tex] => cos A = [tex]\sqrt{1-\frac{(-4)^{2} }{5^{2} } } =\frac{3}{5}[/tex] (because A is in quadrant IV)
tan B = [tex]\frac{5}{12}[/tex]
have:
[tex]1+ cot^{2}B=\frac{1}{sin^{2}B } => 1+\frac{1}{tan^{2} B}=\frac{1}{sin^{2}B } \\=> 1+\frac{1}{\frac{5^{2} }{12^{2} } } =\frac{1}{sin^{2}B }\\ => sin^{2}B=\frac{25}{169}[/tex]
because B is in quadrant III => [tex]sin B=-\sqrt{\frac{25}{169} }=\frac{-5}{13}=>cosB=-\sqrt{1-\frac{5^{2} }{13^{2} } }=\frac{-12}{13}[/tex]
tan(arcsin(-4/5)+arctan(5/12)) = tan( A + B)
but A,B are the 2 angles of a triangle => tan(A + B) = [tex]\frac{sin(A+B)}{cos(A+B)}[/tex]
have: sin(A+B) = sinA.cosB + cosA.sinB = [tex]\frac{-4}{5}.\frac{-12}{13} +\frac{3}{5}.\frac{-5}{13}=\frac{33}{65}[/tex]
cos(A + B) = cosA.cosB - sinA.sinB =[tex]\frac{3}{5}.\frac{-12}{13}-\frac{-4}{5}.\frac{-5}{13}=\frac{-56}{65}[/tex]
=> tan(A + B) = [tex]\frac{33}{65}:\frac{-56}{65}=\frac{-33}{56}[/tex]
