Recall the sum of angles for tan:
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
You have the reciprocal of that, so what you have is
cot(200-80) = cot(120) = -1/√3
how do i evaluate this?
1+tan 200(degrees) tan 80(degrees)/tan200(degrees)-tan 80(degrees)
2 answers
I am sure you meant:
(1+tan 200° tan 80°)/ (tan200°-tan 80°)
--- those brackets are critical
recall:
tan(A-B)
= (tanA - TanB)/( 1 + tanAtanB)
= 1/[ (1+tanAtanB)/(tanA - tanB) ]
tan(200-80) = tan 120 = - tan60° = -√3
= (tan 200 - tan80)/(1 + tan200tan80)
= -1/√3
(1+tan 200° tan 80°)/ (tan200°-tan 80°)
--- those brackets are critical
recall:
tan(A-B)
= (tanA - TanB)/( 1 + tanAtanB)
= 1/[ (1+tanAtanB)/(tanA - tanB) ]
tan(200-80) = tan 120 = - tan60° = -√3
= (tan 200 - tan80)/(1 + tan200tan80)
= -1/√3