Can you tell me how to use the formula Tana = Tan (a + 360 * k) to calculate Tan 600? RT, I don't know how to calculate it when I convert it to tan 240 Isn't k an integer? K on the first floor doesn't take an integer?

tan600=tan(600-360×2)＝tan（-120）＝－tan（120）＝-√3
K is an integer, but integers are also positive and negative
Heat and mass transfer department graduate students for your answer, for reference

tan(2π-a)=-tana If the odd and even are the same, it should be Tana, isn't it a God?

The odd even invariant is for sine and cosine

tanA=2,tan(π/4+A)=

tan(π/4+A)=(tanπ/4+tanA)/(1-tanπ/4tanA)=(1+2)/(1-1*2)=-3
Happy New Year_ Thank you

If √ 3 / 3 is a root of the quadratic equation x ^ 2 - (Tan θ + 1 / Tan θ) x + 1 = 0, Tan θ

√ 3 / 3, so tan θ = √ 3 / 3
Tan2 θ = √ 3 select b

If (radical 3) / 3 is a root of the quadratic equation X2 - (Tan α + 1 / Tan α) x + 1 = 0, Tan α

The original equation can be reduced to x 2 - (Tan α + 1 / Tan α) x + Tan α· (1 / Tan α) = 0
The left factorization shows that: (x - Tan α) (x - 1 / Tan α) = 0
The solution is: x = Tan α or x = 1 / Tan α
Then we know that the two real roots of the equation have the same sign
If (radical 3) / 3 is a root of the equation, then both of the equations are positive numbers
Because Tan α < 1, so: 1 / Tan α > 1
It can be seen that Tan α = (radical 3) / 3
Therefore, tan2 α = 2tan α / (1 - Tan 2 α) = 2 (radical 3) / 3 △ (1 - 1 / 3) = radical 3

Tan squared (45 degrees - Φ / 2) =? Φ is 18 degrees

Tan squared (45 degrees - Φ / 2) = [(1-tan Φ / 2) / (1 + Tan Φ / 2)] ^ 2 = (COS Φ / 2-sin Φ / 2) ^ 2 / (COS Φ / 2 + sin Φ / 2) ^ 2 = (1-2sin Φ / 2cos Φ / 2) / (1 + 2Sin Φ / 2cos Φ / 2) = (1-sin Φ) / (1 + sin Φ) Φ is 18 degree Tan square (45 degrees - Φ / 2) = (1-sin18) / (1 + sin18) = 0.528

Cos square (50 degrees + a) + cos square (40 degrees - a) - Tan (30 degrees - a) Tan (60 degrees + a) 0

Cos square (50 degrees + a) + cos square (40 degrees - a) - Tan (30 degrees - a) Tan (60 degrees + a)
=Sin square (90 degrees - 50 degrees - a) + cos squares (40 degrees - a) - Tan (30 degrees - a) cot (90 degrees - 60 degrees - a)
=Sin square (40 degrees - a) + cos square (40 degrees - a) - Tan (30 degrees - a) cot (30 degrees - a)
=1-1=0

Cos α = cos β cos γ. The square of Tan (α + β) / 2 × Tan (α - β) / 2 = Tan γ / 2 was proved

Left = [sin (a + b) / 2 * sin (a-b) / 2] / [cos (a + b) / 2 * cos (a-b) / 2]. After the sum and difference of the product, we can know that the left = [CoSb cosa] / [cosa + CoSb]. If we bring the known conditions into the left = [1-cosr] / [1 + cosr], we can know that the left = [tanr / 2] ^ 2

(1 + Tan square a) * cos square a = 1

Left = cos? A + tan? ACOS? A
=cos²a+(sin²a/cos²a)cos²a
=cos²a+sin²a
=1 = right
Proof of proposition

Tan quadratic?

1 / cos quadratic-1

Can you tell me how to use the formula Tana = Tan (a + 360 * k) to calculate Tan 600? RT, I don't know how to calculate it when I convert it to tan 240 Isn't k an integer? K on the first floor doesn't take an integer?