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1-tan2°cot62°/tan152°-cot88°=k root 3, then k =? I want process

1-tan2°cot62°/tan152°-cot88°=k root 3, then k =? I want process

Grade:9

2 Answers

Faiz
107 Points
5 years ago
Tan(π-x) = - tan xTan(π/2 - x) = cot xCot(π/2 - x) = tan xUse these 3..... -> 1- tan2°cot(90°-28°) / tan(180°-28°) - cot(90-2) -> 1 - tan2°tan28° / -tan28° - tan2° -> 1 - tan2°tan28° / -(tan 28° + tan2°) -> -1 / tan30° = -√3 this gives k= -1 ans....Also.....Tan x+y = tan x + tan y / 1 - tan x tany....
Suriya Vinodh
28 Points
2 years ago
 
Tan(π-x) = - tan x
Tan(π/2 - x) = cot x
Cot(π/2 - x) = tan x
1- tan2°cot(90°-28°) / tan(180°-28°) - cot(90-2)
-> 1 - tan2°tan28° / -tan28° - tan2°
-> 1 - tan2°tan28° / -(tan 28° + tan2°)
-> -1 / tan30° = -√3 this gives k= -1

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