It is proved that Tan (a + b) / Tana = 3 / 2 when 5sinb = sin (2a + b) is known

It is proved that Tan (a + b) / Tana = 3 / 2 when 5sinb = sin (2a + b) is known

5sinB = sin(2A+B) = sin(A+B +A) = sin(A+B)cosA + cos(A+B)sinA,
sinB = sin(A+B-A) = sin(A+B)cosA - cos(A+B)sinA
5sin(A+B)cosA - 5cos(A+B)sinA = sin(A+B)cosA + cos(A+B)sinA
4sin(A+B)cosA = 6cos(A+B)sinA,
tan(A+B)cotA = 6/4 = 3/2

If 5sinb = sin (2a + b), Tan (a + b) = 9 / 4, then Tana=

5sinB=sin(2A+B)
5sin[(A+B)-A]=sin[(A+B)+A]
5sin(A+B)cosA-5cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA
4sin(A+B)cosA=6cos(A+B)sinA
So,
4tan(A+B)=6tanA
So,
tanA = (2/3)tan(A+B)
=3/2

It is known that 0 < a < π / 4,0 < B < π / 4,5sinb = sin (2a + b), 4tan (A / 2) = 1-tan ^ 2 (A / 2) to find a + B

4tan(a/2)=1-tan^2(a/2)
tana=2tan(a/2)/[1-tan^2(a/2)]=1/2
5sinb=sin(2a+b)
5sin[(a+b)-a]=sin[(a+b)+a]
5sin(a+b)cosa-5cos(a+b)sina=sin(a+b)cosa+cos(a+b)sina
2sin(a+b)cosa=3cos(a+b)sina
tan(a+b)=sin(a+b)/cos(a+b)=(3/2)sina/cosa=(3/2)tana=3/4
So a + B = arctan (3 / 4)

Given that sin (2a + b) = 5sinb, it is proved that 2tan (a = b) = 3tana

5sin [(a + b) - A] = sin [(a + b) + a] left = 5sin (a + b) cosa - 5cos (a + b) Sina, right = sin (a + b) cosa + cos (a + b) Sina, the shift term is: 4sin (a + b) cosa = 6cos (a + b) Sina both sides divided by 2cosacos (a + b), 2tan (a + b) = 3T

tanπ 8-cotπ The value of 8 is______ .

∵tanπ
8-cotπ
8=sin(π
8)
cos(π
8)−cos(π
8)
sin(π
8)=−cos (π
4)
One
2sin (π
4)=-2，
The answer is: - 2

It is proved that (Tan ^ 2A + Tana + 1) (COT ^ 2A + COTA + 1) = Tan ^ 2A + cot ^ 2A + 1

(Tan ^ 2A + Tana + 1) (COT ^ 2A + COTA + 1)
(tan^2a+tana+1)(cot^2a-cota+1)=tan^2a+cot^2a+1
tan^2a+cot^2a+1
=tan^2a+1/tan^2a+1
=(tana+1/tana)^2-1
=(tana+1/tana+1)(tana+1/tana-1)
=(tan^2a+tana+1))[(tana+1/tana-1)/tana]
=(tan^2a+tana+1)(1+1/tan^2a-1/tana)
=(tan^2a+tana+1)(cot^2a-cota+1)

(1 + Tan ^ 2a) / (1 + cot ^ 2a) = (1-tana / 1-cota) ^ 2

(1 + Tan ^ 2a) / (1 + cot ^ 2a) = [(COS ^ 2A + sin ^ 2a) / cos ^ 2A] / [(sin ^ 2A + cos ^ 2a) / sin ^ 2A] / [(sin ^ 2A + cos ^ 2a) / sin ^ 2A] = sin ^ 2A / cos ^ 2A = Tan ^ 2A; (1-tana / 1-cota) ^ 2 = {[(cosa-sina) / cosa] / [(sina-cosa) / Sina]} ^ 2 = (Sina / COSA) ^ 2 = (Sina / COSA) ^ 2 = 2 = Tan ^ 2A. So: (1 + Tan ^ 2a) / (1 + cocos ^ 2a) / sin ^ 2A] = sin ^ 2 = Tan ^ 2A. So: (1 + Tan ^ 2a) / (1 + cos + cos ^ 1 + cos ^ 2) / (1 + cosa / 1 T ^

(tan^2a-cot^2a)/(sin^2a-cod^2a)=sec^2a+csc^2a 2 means that the second power is not together with a

(tan²a-cot²a)/(sin²a-cos²a)=(sin²a/cos²a-cos²a/sin²a)/(sin²a-cos²a)=(sin^4 a-cos^4a)/(sin²a*cos²a)(sin²a-cos²a)=(sin²a-cos
...

(1) ^ 2cot2a (simplified Mathematics)

(2cotA)/(1-cot²A)
=(2/tanA)/(1-1/tan²A)
=2tanA/(tan²A-1)
=-2tanA/(1-tan²A)
=-tan2A
If you don't understand, I wish you a happy study!

It is known that the moving circle x2 + y2-2mx-4my + 6m-2 = 0 is constant over a fixed point, and the coordinates of this fixed point are______ .

x2+y2-2mx-4my+6m-2=0，
∴x2+y2-2=（2x+4y-6）m，
Qi
x2+y2−2＝0
2x+4y−6＝0 ，
The solution is x = 1, y = 1, or x = 1
5，y=7
5，
The coordinates of the fixed point are (1, 1), or (1)
5，7
5）．
So the answer is: (1,1), or (1)
5，7
5）．