tanθ·tan2θ+tan2θ·tan3θ+.+tannθ·tan(n+1)θ To complete the process

tanθ·tan2θ+tan2θ·tan3θ+.+tannθ·tan(n+1)θ To complete the process

tanαtanβ+1=（tanα-tanβ）/tan（α-β）tanθtan2θ+1=(tan2θ-tanθ)/tan(2θ-θ)=(tan2θ-tanθ)/tanθtan2θtan3θ+1=(tan3θ-tan2θ)/tan(3θ-2θ)=(tan3θ-tan2θ)/tanθ.tan(nθ)tan(n+1)θ+1=[tan(n+1)θ...

What's three out of two

Cos2 3 / 3 = 0

If Tan (π 4 − θ) = 3, then Cos2 θ 1 + sin2 θ=______ ．

∵ Tan (π 4 - θ) = 1 − Tan θ 1 + Tan θ = 3, ∵ Tan θ = - 12, then Cos2 θ 1 + sin2 θ = Cos2 θ − sin2 θ sin2 θ + Cos2 θ + 2Sin θ cos θ = 1 − tan2 θ tan2 θ + 1 + 2tan θ = 1 − 1414 + 1 − 1 = 3

In △ ABC, we know that (a + B + C) (a + B-C) = 3AB, sinasinb = 3-4, and judge the shape of △ ABC

What's in gold

In △ ABC, the equation that must hold is ()
A. asinA=bsinBB. acosA=bcosBC. asinB=bsinAD. acosB=bcosA

According to the sine theorem: Asina = bsinb, that is, asinb = bsina

If the three internal angles of a triangle are ABC, then the following equation must be true:
A.sin（A+B）=-sinA
B.cos((B+C)/2)=sin(A/2)
C.tan((B+C)/2)=tan(A/2)
D.sin((B+C)/2)=-sinA/2

A+B+C=π
(B+C)/2=(π-A)/2≤π/2;
(B + C) / 2 is the acute angle;
So option B
cos((B+C)/2)=sin(A/2)
establish

As shown in the figure, in △ ABC, if ∠ ade = ∠ B is known, then the following equation holds ()
A. ADAB＝AEACB. AEBC＝ADBDC. DEBC＝AEABD. DEBC＝ADAC

∵∵∵∵∵ a = ∵ a, ∵ ade = ∵ B, ∵△ AED ∵ ACB, ∵ aeac = ADAB

Given Tan θ = 2. (I) find the value of Tan (π 4 + θ); (II) find the value of COS 2 θ

(I) ∵ Tan θ = 2, ∵ Tan (π 4 + θ) = 1 + Tan θ 1 − Tan θ = 1 + 21 − 2 = - 3; (II) ∵ Tan θ = 2, ∵ sin θ cos θ = 2, that is sin θ = 2cos θ, ∵ sin2 θ = 4cos2 θ, ∵ 1-cos2 θ = 4cos2 θ, that is Cos2 θ = 15,