Sec²A(1+cot²A) =1
from trig identity
sin²A + cos²A = 1 (dividing thru by sin²A)
=> sec²A = 1 + tan²A and cot²A = 1/tan²A
Substituting the above into the equation
(1 + tan²A)(1 + 1/tan²A) = 1
Multiplying through by tan²A, it will be
(1 + tan²A)(1 + tan²A) = tan²A
tan⁴A + 2tan²A + 1 = tan²A
tan⁴A + tan²A + 1 = 0
Let x = tan²A
=> x² + x + 1 = 0
Solving using quadratic formula
=> x = ½(-1 + i√3) or ½(-1 - i√3)
So, tanA = + √(½(-1 + i√3)) or - √(½(-1 + i√3))
Or tanA = + √(½(-1 - i√3)) or - √(½(-1 - i√3))
Then, A = tan-¹(+√(½(-1 + i√3))) or
A = tan-¹(-√(½(-1 + i√3))) or
A = tan-¹(+√(½(-1 - i√3))) or
A = tan-¹(-√(½(-1 - i√3)))
3 months ago
from trig identity
sin²A + cos²A = 1 (dividing thru by sin²A)
=> sec²A = 1 + tan²A and cot²A = 1/tan²A
Substituting the above into the equation
(1 + tan²A)(1 + 1/tan²A) = 1
Multiplying through by tan²A, it will be
(1 + tan²A)(1 + tan²A) = tan²A
tan⁴A + 2tan²A + 1 = tan²A
tan⁴A + tan²A + 1 = 0
Let x = tan²A
=> x² + x + 1 = 0
Solving using quadratic formula
=> x = ½(-1 + i√3) or ½(-1 - i√3)
So, tanA = + √(½(-1 + i√3)) or - √(½(-1 + i√3))
Or tanA = + √(½(-1 - i√3)) or - √(½(-1 - i√3))
Then, A = tan-¹(+√(½(-1 + i√3))) or
A = tan-¹(-√(½(-1 + i√3))) or
A = tan-¹(+√(½(-1 - i√3))) or
A = tan-¹(-√(½(-1 - i√3)))