tan A+ tan (60+A) -tan (60-A) sama dengan (1) 3 tan 3A (2) tan 3A (3) cot 3A (4) sin 3A

tan A+ tan (60+A) -tan (60-A) sama dengan (1) 3 tan 3A (2) tan 3A (3) cot 3A (4) sin 3A

Jawaban 1)

Mari kita hitung tan (60 + A) dan tan (60 – A).

(tan left ( 60 + A right ) = frac{tan 60^{0}+ tan A}{1 – tan60^{0} tanA})

(tan left ( 60 + A right ) = frac{sqrt{3}+tan A}{1 – sqrt{3} tanA})…………………….(1)

(tan left ( 60 – A right ) = frac{tan 60^{0}- tan A}{1 + tan60^{0} tanA})

(tan left ( 60 – A kanan ) = frac{sqrt{3} – tan A}{1 + sqrt{3} tanA})……………………..(2

Mengurangi persamaan (1) dan (2)

(tan left ( 60 + A right ) – tan left ( 60 – A right ) = frac{sqrt{3}+tan A}{1 – sqrt{3} tanA} – frac{ sqrt{3} – tan A}{1 + sqrt{3} tanA}) (tan left ( 60 + A right ) – tan left ( 60 – A right ) = frac{left ( sqrt{3} + tan A right)left ( 1+ sqrt{3}tan A right) – left ( sqrt{3} – tan A right)left ( 1- sqrt{ 3}tan A right )}{1^{2} – left ( sqrt{3} tan A right )^{2}}) (tan left ( 60 + A right ) – tan kiri ( 60 – A kanan ) = frac{sqrt{3} + 3tan A + tan A + sqrt{3} tan^{2}A – sqrt{3} + 3 tan A + tan A – sqrt{3} tan^{2}A}{1 – 3 tan^{2}A}) (tan left ( 60 + A right ) – tan left ( 60 – A right ) = frac {8 tan A}{1 – 3 tan^{2}A}) (tan A + tan left ( 60 + A right ) – tan left ( 60 – A right ) = tan A + frac {8 tan A}{1 – 3 tan^{2}A}) (tan A + tan left ( 60 + A right ) – tan left ( 60 – A right ) = frac{tan A – 3 tan^{3}A + 8 tan A}{1 – 3 tan^{2}A}) (tan A + tan left ( 60 + A right ) – tan left ( 60 – A kanan ) = frac{9 tan A – 3 t an^{2}A}{1 – 3 tan^{2}A}) (tan A + tan left ( 60 + A right ) – tan left ( 60 – A right ) = frac{ 3left ( 3 tan A – tan^{3}A kanan )}{1 – 3 tan^{2}A}) (tan A + tan left ( 60 + A right ) – tan left ( 60 – A kanan ) = 3 tan 3A)

Di sini tan 3A = (tan 3A = frac{3 tan A – tan^{3}A}{1 – 3 tan^{2}A})

10
Author: fungsi