5. 1
Given that x
tan y 2 and then
tan x
2
tan y
Find !

6. tan (x y) 2 tan x tan y
tan x y
1 tan x.tan y
1
tan x
2 tan x tan y
2
1 tan x. tan y
2 2 tan x.tan y tan x tan y
1 1
2 2. .tan y tan y
2 2
1
tan y .....? 2 tan y tan y
2
1
2 2 tan y
2

7. 1
2 2 tan y
2
3
2 tan y
2
3 1
tan y .
2 2
3
tan y
4
3
So, the value of tan y is equal
4

8. Prove that
1
cosec x cot x cot x
2

9. 1
cosec x cotx cot x
2
1 1 1
cot x . x
2 tan 2
1
sin x
1 cos x
1 cos x
sin x

10. 1 cos x
sin x
1 cos x
sin x sin x
cosec x cot x
1
cot x cosec x cotx
2
1
cosec x cot x cot x
2

11. o o
cos 1 sin
Let f( ) o o
1 sin cos
a.) Prove that f( ) 2 sec o
b.) Solve the equation f 4,
giving your answer for A in
the interval 0 360o

12. o
cos o
1 sin o f( ) 2 sec
f( ) o o
1 sin cos
cos Α o 1 sin Α o cos 2 A o 2sin A o sin 2 A o
f (A)
1 sin Α o cos Α o 1 sin A o . cos A o
cos 2 A o sin 2 A o 1 2sin A o
1 sin A o . cos A o
1 1 2sin A o
1 sin A o . cos A o

13. 1 1 2sin A o
1 sin A o . cos A o
2 2sin A o
1 sin A o . cos A o
2(1 sin A o )
1 sin A o . cos A o
2
cos A o
2 sec A o
cos Α o 1 sin Α o
f (A) 2 sec x
1 sin Α o cos Α o

14. f 4
0 360o
f( ) 2 sec o 2sec A o 4
sec A o 2
f 4 1
2
cos A o
1
cos A o
2
cos A o cos 60o

15. cos A o cos 60o in the interval 0 360o
x α o k.360o
A 60o k.360o
k 0 A 60o
x α o k.360o
A 60o k.360o
k 0 A 300o
So, the value A in the interval 0 360o
are 60o and 300o

16. Evaluate each of the following :
a. tan180o 2. cos180o 3. csc270o sin 90o
b. sin 0o 3. cot 90o 5.sec180o 4. cos270o

17. a. tan180o 2. cos180o 3. csc270o sin 90o
0 2. 1 3. 1 1
2 3 1
0
So,
o o o o
tan180 2. cos180 3. csc270 sin 90 0

18. b. sin 0o 3. cot 90o 5. sec180o 4 cos 270o
0 3. 0 5. 1 40
0 0 5 0
5
So,
o o o o
sin 0 3. cot 90 5. sec180 4 cos270 5

19. Prove that
1 cos x sec x 1 2
cot x cosec x
1 cos x sec x 1

20. 1 cos x sec x 1 2
cot x cosec x
1 cos x sec x 1
1 cos x sec x 1 Prove the right side
1 cos x sec x 1
1 cos x
sec x 1 cos x cos x
sec x 1 1 cos x
cos x cos x
1 cos x
cos x
1 cos x
cos x

21. 1 cos x
cos x
1 cos x
cos x
1 cos x cos x
.
cos x 1 cos x
1 cos x
1 cos x
1 cos x sec x 1
1 cos x sec x 1

22. 1 cos x sec x 1 2
cot x cosec x
1 cos x sec x 1
sec x 1 2
cot x cosec x Prove the right side
sec x 1
2
cot x cosec x cot 2 x cosec2 x 2 cot x.cose cx
cos 2 x 1 cos x 1
2 .
sin 2 x sin 2 x sin x sin x
cos 2 x 1 cos x
2
sin 2 x sin 2 x sin 2 x
cos 2 x 1 2 cos x
sin 2 x

23. cos 2 x 1 2 cos x
sin 2 x
1 sec2 x sec x
2
sec2 x 2
sec x sec2 x
sin 2 x
1 sec2 x 2sec x
sec2 x
1 - cos2 x
1 sec2 x 2 sec x
sec2 x
sec2 x 1
sec2 x sec2 x

24. 1 sec2 x 2 sec x
sec2 x
sec2 x 1
sec2 x sec2 x
1 sec2 x 2 sec x sec2 x
sec2 x sec2 x 1
1 sec2 x 2 sec x
sec2 x 1
(sec x 1) sec x 1
sec x 1 sec x 1
sec x 1
sec x 1
2 sec x 1
cot x cosec x
sec x 1

25. From the first step we sec x 1 1 cos x
could prove that sec x 1 1 cos x
1 cos x sec x 1
1 cos x sec x 1
And from the second 2 sec x 1
cot x csc x
step we could prove sec x 1
that
sec x 1 2
cot x cosec x
sec x 1
So, we can conclude that the statement
1 cos x sec x 1 2
cot x cosec x
1 cos x sec x 1

26. Prove that
4 4
sec x tan x
2 2
1
sec x tan x

27. sec4 x tan 4 x
1
sec2 x tan 2 x Prove the left side
sec4 x tan 4 x sec2 x tan 2 x sec2 x tan 2 x
sec2 x tan 2 x sec2 x tan 2 x
sec2 x tan 2 x
1 sin 2 x
cos 2 x cos 2 x
1 sin 2 x
cos 2 x
cos2 x
cos2 x
1
sec4 x tan 4 x
1
sec2 x tan 2 x

28. In a study of AC circuits, the equation
cos s. cos t
R sometimes arises,
c sin s t
Use a sum identity and algebra to
show this equation is equivalent to
1
R
c tan s tan t

29. cos s. cos t 1
R
c sin s t c tan s tan t
cos s. cos t 1
c sin s t c tan s tan t
cos s. cos t 1 cos s. cos t
.
c sin s t c (sin s. cos t sin t. cos t )
1 1
.
c (sin s. cost sin t. cost )
cos s. cost

30. 1 1
.
c (sin s. cost sin t. cost )
cos s. cost
1
sin s. cos t sin t. cos s
c
cos s. cos t cos s. cos t
1
sin s sin t
c
cos s cos t
1
c tan s tan t
cos s. cos t 1
c sin s t c tan s tan t

32. Show that the circle 2
x2 y 6x 2 y 54 0
x2 and22 x 8y 112 0
y2
are tangent
internally
(without a graphing)

33. x2 y2 6x 2 y 54 0 …….I
x2 y 2 22 x 8y 112 0 …….Ii
28x 6y 166 0
14x 3y 83 0
3 y 83 14 x
83 14 x
y
3
Then substitution to the first equation
x2 y2 6x 2 y 54 0
2
2 83 14 x 83 14 x
x 6x 2 54 0
3 3
6889 2324x 196x 2 166 28x
x2 6x 54 0
9 3 3

34. 2 6889 2324x 196x 2 166 28x
x 6x 54 0
9 3 3
205 2 6889 498 486 2324x 54 x 84 x
x 0
9 9 9
205 x 2 (6889 498 486 ) 2324 x 54 x 84 x 0
205x 2 2186x 5905 0
205x 2 of
Then the value for the discriminant 2186x 5905 0
D b2 4.a.c
2
2186 4.205 .5905
477856 4842100
4364244
D 0 The circle do not intersect

35. Find the equation of the circle
through the point (5,1), (4,6) and
(2,-2)

36. Trough Point
(5,1) x2 y2 Ax By C 0
2
52 1 5A 1B C 0
25 1 5A B C 0
5A B C 26 …….Iii
Trough Point
(4,6) x 2 y 2 Ax By C 0
2
42 6 4A 6B C 0
16 36 4A 6B C 0
4A 6B C 52
2A 3B C 26 …….Iii

37. Trough Point
(2,- x2 y2 Ax By C 0
2
2) 22 2 2A 2B C 0
4 4 2A 2B C 0
2A 2B C 8
A B C 4 …….Iii
Then, from I, and III equation we get . .
.
And from II, and III equation we get . . .
5A B C 26 …….I
A B C 4 …….Iii 2A 3B C 26 …….Ii
6A 2B 22 A B C 4 …….Iii
3A B 11 …….Iv
A 4B 22 …….v

38. From elimination IV and V equation, we can get the value of A
3A B 11 …….Iv 4 12A 4B 44
A 4B 22 …….v 1 A 4B 22
11A 22
A 2
Then, we are going to find the
value of B from IV equation And the value of C can we get by
3A B 11 …….Iv substitution of the III equation
3 2 B 11 A B C 4
6 B 11 2 5 C 4
B 5 C 1

39. After that, substitution the value of A, B,
and C that we already got to the
equation ofBy Ccircle
2 2
x y Ax
the 0
2 2
A 2
x y 2x 5y C 0
B 5
C 1
2 2
So, the equation of the x y is x
circle 2 5y C 0

40. a. Determine the equation of a circle that
passes through points (-2,4) and x(7,7) 7
y
and has a center on line
b. Determine the equation of tangent line
at points (7,7) on the circle
c. Determine the equation of the tangent
x 2y 1 0
line on a circle which is parallel to line

41. Given
Aske
thatpoint A = (-2,4)
Trough : d:
Center on
and B(7,7)line x+y=7 The equation of ci
Answe
r:
Suppose that the equation of
x 2 circle isBy C 0
the y 2 Ax
Trough 2point A (-2,4)
2
x y Ax By C 0
4 16 2A 4B C 0
2A 4B C 20 …... i

42. Trough point A (-2,4)
x2 y2 Ax By C 0
49 49 7A 7B C 0
7A 7B C 98 …... ii
The center of
circle 1 A, 1 B
C
2 2
x y
Then, substitution the center of the circle to the equation of
center line circle
x y 7 x y 7
1 1
A B 7
2 2
A B 14 …... iii

43. Then, elimination the I and the II
equation
2A 4B C 20 …... i
7A 7B C 98 …... ii
9A 3B 78
3A B 26 …... iV
And from elimination Iv
and Iii equation we can
get the value of A
3A B 26 …... iv Then, the value of B can
A B 14 …... iii we get from Iii equation
A B 14 iii
…...
2A 12
6 B 14
A 6
B 8

44. The value of A and B are known, so we can get the value
C fromAequation 20 …... i
2 I 4B C
2 6 4 8 C 20
12 32 C 20
C 0
After that, substitution the value of A, B, and C
that we already got to the equation of the
circle y 2 Ax By C 0
x2 A 6
x2 y2 6x 8y 0 B 8
C 0
So, the equation of
circle is x 2 y 2 6x 8y 0

45. Given
Aske
that :
The equation of the circle d:
The equation of tangent
x2 y2 6x 8y 0
line at points (7,7) on the
circle
Answe
r:
x2 y2 6x 8y 0
72 72 67 87 0
49 49 42 56 0
0 0 on the circle

46. The equation of
1
tangent line(x1 x) 1
x1x y1 y A is B (y1 y) C 0
2 2
1 1
7x 7 y 6 (7 x) 8 (7 y) 0 0
2 2
7x 7 y 3(7 x) 4(7 y) 0 0
7x 7 y 21 3x 28 4y 0
4x 3y 49 0
So, the equation of tangent line is
4x 3y 49 0

47. Given Aske
that :
The equation of the circle
d:
the equation of the
x2 y2 6x 8y 0 tangent line on a circle
x 2y 1 0
which is parallel to line
Answe
r: x 2y 1
Find the gradient of the equation 0
x 2y 1 0
2y x 1
1 1 1
y x So,
2 2 2
m=

48. Find the center and radius of the circle x 2 has
which y2 equation 0
6x 8y
1 2 1 2 1 1
r A B C C A, B
4 4 2 2
1 2 1 1 1
6 ( 8) 2 0 C 6, ( 8)
4 4 2 2
9 16
C(3,4)
So, the center is
25
(3,4)
5 So, r =
5

49. The equation of the tangent line L a2circle 6x 8y 0
on x y 2
which is parallel toxline 1 0
2y
2 1
y b m( x a) r 1 m m
2
2
1 1 r 5
y 4 (x 3 ) 5 1
2 2
C(3,4)
1 3 1
y 4 x 5 1
2 2 4
a, b
1 3 5
y x 4 5 0
2 2 4
1 5 5
y x 5 0
2 2 2
2y x 5 5 5 0

50. 2y x 5 5 5 0
2y x 5 5 5 0 And 2y x 5 5 5 0
The equation of the tangent L a 2circle 6x
line on x y2 8y 0
which is parallel to x 2y 1 0
line are
2y x 5 5 5 0 And 2y x 5 5 5 0