Use trigonometric identities to simplify the expression. 1/cot^2(B) +1.....the B represents the symbol of beta.
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
we know that
The equation of a circle in standard form is equal to
In this problem we have
substitute
Answer:
(a) The solution is x=47.
(b) The solution is x=223755.
(c) The solution is x=33703314.
(d) The solution is x=984414.
Step-by-step explanation:
(a) Step 1 is to solve
q e
2 4 265 250 Calculation I
3 3 374 335 Calculation II
Now Solving for calculation I:
x≡≡
Solve (265)x=250(mod 433) for x0,x1,x2,x3.
x0:(26523)x0=25023(mod 433)⟹(432)x0=432⟹x0=1
x1:(26523)x1=(250×265−x0)22(mod 433)=(250×265−1)22(mod433)=(250×250)22(mod 433)⟹(432)x1=432⟹x1=1
x2:(26523)x2=(250×265−x0−2x1)21(mod 433)=(250×265−3)22(mod 433)=(250×195)21(mod 433)⟹(432)x2=432⟹x2=1
x3:(26523)x3=(250×265−x0−2x1−4x2)20(mod 433)=(250×265−7)20(mod 433)=(250×168)20(mod 433)⟹(432)x3=432⟹x3=1
Thus, our first result is:
x≡x0+2x1+4x2+8x3(mod24)≡1+2+4+8(mod24)≡15(mod24)
Now for Calculation II:
x≡≡
Solve (374)x=335(mod 433) for x0,x1,x2.
x0:(37432)x0=33532(mod 433)⟹(234)x0=198⟹x0=2. Note: you only needed to test x0=0,1,2, so it is clear which one x0 is.
x1:(37432)x1=(335×374−x0)31(mod 433)=(335×374−2)31(mod 433)=(335×51)31(mod 433)=1(mod 433)⟹(234)x1=1(mod 433)⟹x1=0
x2:(37432)x2=(335×374−x0−3x1)30(mod 433)=(335×374−2)30(mod 433)=(335×51)30(mod 433)=198(mod 433)⟹(234)x2=198(mod 433)⟹x2=2. Note: you only needed to test x2=0,1,2, so it is clear which one x2 is.
Thus, our second result is:
x≡x0+3x1+9x2(mod 33)≡2+0+9×2(mod 33)≡20(mod 33)
Step 2 is to solve
x ≡15 (mod 24 ),
x ≡20 (mod 33 ).
The solution is x=47.
(b) Step 1 is to solve
q e
2 10 4168 38277 523
3 6 674719 322735 681
is calculated using same steps as in part(a).
Step 2 is to solve
x ≡ 523 (mod 210 ),
x ≡ 681 (mod 36 ).
The solution is x=223755 .
(c) Step 1 is to solve
q e
2 1 41022298 1 0
29 5 4 11844727 13192165
In order to solve the discrete logarithm problem modulo 295 , it is best to solve it step by step. Note that 429 = 18794375 is an element of order 29 in F∗p . To avoid notational confusion, we use the letter u for the exponents.
¢294
First solve 18794375u0 = 11844727
= 987085.
The solution is u0 = 7.
The value of u so far is u = 7.
¢293
Solve 18794375u1 = 11844727·4−7
= 8303208.
The solution is u1 = 8.
The value of u so far is u = 239 = 7 + 8 · 29.
¢292
Solve 18794375u2 = 11844727 · 4−239
= 30789520.
The solution is
u2 = 26. The value of u so far is u = 22105 = 7 + 8 · 29 + 26 · 292 .
¢291
Solve 18794375u3 = 11844727 · 4−22105
= 585477.
The solution is
u3 = 18. The value of u so far is u = 461107 = 7 + 8 · 29 + 26 · 292 + 18 · 293 .
¢290
Solve 18794375u4 = 11844727 · 4−461107
= 585477.
The solution is
u4 = 18. The final value of u is u = 13192165 = 7 + 8 · 29 + 26 · 292 + 18 · 293 + 18 · 294 , which is the number you see in the last column of the table.
Step 2 is to solve
x ≡ 13192165 (mod 295 ).
x ≡ 0 (mod 2),
The solution is x=33703314 .
(d) Step 1 is to solve
q e
2 1 1291798 1 0
709 1 679773 566657 322
911 1 329472 898549 534
To solve the DLP’s modulo 709 or 911, they can be easily solved by an exhaustive search on a computer, and a collision algorithm is even faster.
Step 2 is to solve
x ≡ 0 (mod 2),
x ≡ 322 (mod 709),
x ≡ 534 (mod 911).
The solution is x=984414
let's firstly convert the mixed fractions to improper fractions and then sum them up.