In the triangle ABC, what should we do to prove Tan (A / 2) + Tan (B / 2) - cot (C / 2) = - Tan (A / 2) Tan (B / 2) cot (C / 2)? In the triangle ABC, we prove that Tan (A / 2) + Tan (B / 2) - cot (C / 2) = - Tan (A / 2) Tan (B / 2) cot (C / 2)

In the triangle ABC, what should we do to prove Tan (A / 2) + Tan (B / 2) - cot (C / 2) = - Tan (A / 2) Tan (B / 2) cot (C / 2)? In the triangle ABC, we prove that Tan (A / 2) + Tan (B / 2) - cot (C / 2) = - Tan (A / 2) Tan (B / 2) cot (C / 2)

tan(A/2)+tan(B/2)-cot(C/2)
=tan(A/2)+tan(B/2)-tan(π/2-C/2)
=tan((A+B)/2)(1-tan(A/2)tan(B/2))-tan((A+B)/2)
=tan((A+B)/2)-tan((A+B)/2)tan(A/2)tan(B/2)-tan((A+B)/2)
=-tan((A+B)/2)tan(A/2)tan(B/2)
=-tan(A/2)tan(B/2)tan(π/2-C/2)
=-tan(A/2)tan(B/2)cot(C/2)

In acute triangle ABC: simplify Tan (A / 2) Tan (B / 2) + Tan (B / 2) Tan (C / 2) + Tan (C / 2) Tan (A / 2)

Original formula = Tan (A / 2) * + Tan (B / 2) * Tan (C / 2)
=tan(A/2)*tan*+tan(B/2)*tan(C/2)
=tan(A/2)*cot(A/2)*+tan(B/2)*tan(C/2)
=1
Application formula: Tana + tanb = Tan (a + b) * (1-tana * tanb)
I don't know how to type the brackets, but I use ﹥ instead

In the triangle ABC, M is the midpoint of B and C, the three sides of the triangle AMC are three consecutive integers, and Tanc = cotbam
In the triangle ABC, M is the midpoint of BC, the three sides of ABC are three consecutive integers, and Tanc = cot ∠ BAM
(1) Judging the shape of a triangle
(2) Find cosa

(1) Let BAM = α and MAC = β,
Then, from Tanc = cot α, α + C = 90 °, β + B = 90 °
In △ ABM, from the sine theorem, bmsin α = amsinb, that is sinbsin α = ammb
Similarly, sincsin β = ammc,
∵MB=MC,∴sinBsinα=sinCsinβ,
∴sinαsinC=sinβsinB∵α+C=90°,β+B=90°,∴sinαcosα=sinβcosβ
That is, sin2 α = sin2 β, ∧ α = β or α + β = 90 degree
When α + β = 90 °, am = 12bc = MC,
It is contradictory to the fact that the length of three sides of △ AMC is three consecutive positive integers,
The results show that the triangles are isosceles
(2) In right triangle AMC, let two right sides be n, n-1 and the hypotenuse n + 1 respectively,
From (n + 1) 2 = N2 + (n-1) 2, n = 4,
From cosine theorem or double angle formula, cos ∠ BAC = 7 / 25
Or cos ∠ BAC = - 7 / 25

In △ ABC, point m is the midpoint of BC side, the side length of △ AMC is three consecutive positive integers, and Tan angle c = cot angle BAM
1 judge the shape of △ ABC 2 find the cosine value of the angle BAC. Everyone, I'm in a hurry. Help me. If the process is more comprehensive, I'll add points

(1) From the fact that the side length of △ AMC is three consecutive positive integers, we can associate with the law of "three strands, four chords and five chords". It can be concluded that the side lengths of △ AMC are 3, 4 and 5 respectively, which are right triangles, and the right angle is ∠ AMC

If a, B and C are three sides of △ ABC, the result of simplifying | a-b-c | + | b-c-a | + | C-A-B |, is ()
A. -a-b-cB. a+b+cC. a+b-cD. a-b+c

∵ a, B, C are the three sides of △ ABC, ∵ a < B + C, B < C + A, C < A + B, ∵ a-b-c < 0, b-c-a < 0, C-A-B < 0, ∵ a-b-c | + | b-c-a | + | C-A-B | = B + C-A + C + A-B + A + B-C = a + B + C

Sin = contrast slant cos = neighbor ratio slant Tan = contrast slant which side is the opposite side, the hypotenuse and the adjacent side in the right triangle?

I'm looking for the angle

Sin = contrast slant cos = neighbor ratio slant Tan = contrast slant which side is the opposite side, the hypotenuse and the adjacent side in the right triangle?

If there is a point P on the upper support of hyperbola Y & # 178 / 12 & # 178; - X & # 178 / 5 & # 178; = 1 and the distance from the focus is equal to 3, then the coordinate of point P is equal to 3——————
I'm not right=

Let P (x ', y')
It is known that a = 12, B = 5, C = 13, e = 13 / 12
Then ey '- 12 = (13 / 12) y' - 12 = 3 (focal radius formula)
y'=180/13
(180/13)^2/12^2-x‘^2/5^2=1
We get X '= - 10 (√ 14) / 13 or X' = 10 (√ 14) / 13
So p (- 10 (√ 14) / 13180 / 13) or P (10 (√ 14) / 13180 / 13)
Hope to help you!

Hyperbola X & # 178; - Y & # 178; = 1, the distance from a point P to the left quasilinear is 1, then the distance from point P to the right focus is? Detailed process

The distance from a point P on the hyperbola X & # 178; - Y & # 178; = 1 to the left quasilinear is 1,
Then the distance from P to the left focus D1 = root 2
Then the distance from point P to the right focus is root 2 + 2

It is known that 2sin2 α + 2Sin α cos α 1 + Tan α = K (0 ＜ α＜ π 2)

2sin2 α + 2Sin α cos α 1 + Tan α = 2Sin α (sin α + cos α) 1 + sin α cos α = 2Sin α cos α (sin α + cos α) sin α + cos α = 2Sin α cos α = K