1. ( 11.Trigonometry )
Trigonometry is derived from Greek words tri ( three) gonon (angles) and metron ( measure).
Trigonometry means the measure of three angles in a right triangle .
Trigonometry specifically deals with the relationships between the sides and the angles of triangles, that is, on the trigonometric functions, and with calculations based on these functions.

2. Relations between sides and angles of a right triangle
If an angle of a triangle contains 900 , then that triangle said to be right angle triangle.
Opposite side
hypotenuse
An angle which contains 900 then that angle is said to be right angle .
Adjecent Side
The side which is opposite to right angle is called hypotenuse .It denote by AB or c .
The side opposite to an angle  is called opposite side to and it is denoted by BC or a .
The side adjacent to an angle  is called adjacent side to  and it is denoted by AC or b .

3. Ratioes of a right triangle.
The ratio of opposite side and hypotenuse is called Sine A . Shortly it is written as Sin A and read as Sine A.
Opposite side to A / hypotenuse
Opposite side
hypotenuse
Adjacent side
The ratio of adjacent side and hypotenuse is called Cosine A . Shortly it is written as Cos A and read as Cos A
adjacent side to A / hypotenuse

4. = Opposite side to A / adjacent side to A hypotenuse
The ratio of Opposite side and adjacent side is called Tangent A . Shortly it is written as Tan A and read as Tan A
Opposite side
= Opposite side to A / adjacent side to A
hypotenuse
Adjacent side
The ratio of adjacent side and Opposite side is called Cotangent A . Shortly it is written as Cot A and read as Cot A
adjacent side to A / Opposite side to A

5. = hypotenuse / adjacent side to A
The ratio of hypotenuse and adjacent side is called Secant A . Shortly it is written as Sec A and read as Sec A
= hypotenuse / adjacent side to A
Opposite Side
hypotenuse
Adjacent Side
The ratio of hypotenuse and Opposite side is called Secant A . Shortly it is written as Cosec A or csc A and read as Cosec A
= hypotenuse / Opposite side to A

6. Do this
1. Identify hypotenuse”, “Opposite Side” and “adjacent side” for angle R in the given triangle.
Opposite Sice
hypotenuse
for angle R in the adjacent triangle.
PR = hypotenuse
Adjacent Side
PQ = Opposite Side Z
QR = adjacent Side
Opposite Side
2 (i) Identify hypotenuse”, “Opposite Side” and “adjacent side” for angle X in the given triangle.
Adjacent side Y
hypotenuse X
for angle X in the adjacent triangle.
XY = hypotenuse
YZ = Opposite side
XZ =adjacent side

7. Do this
2 (i) Identify hypotenuse”, “Opposite Side” and “adjacent side” for angle Y in the given triangle. Z
Adjacent side
OOpposite side
for angle Y in the adjacent triangle. Y X
hypotenuse
XY = hypotenuse
XZ = Opposite Side
YZ = adjecent Side

8. C {ç³Äæý$†²…^èþ…yìþ: A B C B A hypotenuse
Write lengths of “hypotenuse”, “Opposite Side” and “adjacent side” for the given angles in the given triangle.
1. For angle C For angle A
Adjacent Side
hypotenuse
4 cm
5 cm
According Pythagoras Theorem A B
Opposite side
3 cm
( i ) For angle C
AC = hypotenuse = 5 cm
AB = Opposite Side = 3 cm C
BC = adjacent Side = 4 cm
hypotenuse
( ii ) For angle A
Opposite Side
4 cm
5 cm
AC = hypotenuse = 5 cm
BC = Opposite Side = 4 cm B A
Adjacent Side
AB =adjacent side = 3 cm
3 cm

9. Do This
1. Find (i) Sin C (ii) Cos C and (iii) tan C in the adjacent triangle.
13 cm
5 cm
In the adjacent Δ ABC , B = 900
12 cm
According Pythagoras Theorem cm

10. Do this X
In triangle XYZ , Y is right angle. XZ= 17cm and YZ=15 cm then find (i) sin X (ii) Cos Z (iii ) tan X .
17 cm
8 cm
In triangle XYZ , Y is right angle.
According Pythagoras Theorem Z
15 cm Y
XY = 8 cm

11. Do This 25 cm 24 cm P In triangle PQR with right angle at Q ,
the value of P is x , PQ = 7 cm and QR = 24 cm, then find Sin x and Cos x .
25 cm
7 cm
In triangle PQR with right angle at Q
According Pythagoras Theorem R
24 cm Q
PR = 25 cm

12. Try This A
In a right angle triangle ABC , right angle is at C . BC+CA= 23 cm and BC – CA = 7 cm, then find Sin A and Tan B .
17cm
8 cm
In a right angle triangle ABC , right angle is at C .
15 cm B C
In ΔABC , C = 900
According Pythagoras Theorem
AB = 17 cm

13. Think - Discuss (i) deos exists for some value of angle x ?
Sin value exists always less than 1.
Value is more than 1. So it does not exists for some
value of angle x .
(ii) The Value of Sin A and Cos A is always less than 1. Why?
Origin is the centre of the circle and radius is 1 unit. P(a,b) is the point on the circle. The point is making an angle  with A .
Y Coordinate
X Coordinate

14. According to the Ordered Pairs
P is making an angle at centre to rotate one complete revolution. In adjacent figure AOB is the one fourth part of revolution and making an angle of 900. AOC is the half part of revolution which makes an angle of 1800 at centre of the circle. AOD is the three fourth part of revolution which makes 2700 at the centre.
A,B,C,D points are the ordered pairs which are (1,0),((0,1),(-1,0),(0,-1) respectively .
According to the Ordered Pairs
Sin 00 = Cos 00 = 1
Sin 900 = Cos 900 = 0
Sin 1800 = Cos 1800 = – 1
Sin 2700 = – Cos 2700 = 0
Sin 3600 = Cos 3600 = 1
Like this Sine and Cosine Values are always exists less than 1.
(iii) Tan A is the product of tan and A .
Tan A means Value of the Tangent of an angle A . But not the product of tan and A .

15. Multiplicative Inverses of Trigonometrical Ratioes
Opposite side of A / hypotenuse
hypotenuse
Opposite Side
= hypotenuse / Opposite side of A
Adjacent Side
So Sine and Cosec are called multiplicative inverse trigonometrical ratioes of each other .

16. Multiplicative Inverses of Trigonometrical Ratioes
Adjacent side of A / hypotenuse
Opposite Side
= hypotenuse / Adjacent side of A
hypotenuse
Adjacent Side
So Cosine (Cos) and Secant (Sec) are called multiplicative inverse trigonometrical ratioes of each other .

17. Multiplicative Inverses of Trigonometrical Ratioes
= Opposite side of A / adjacent side of A
Opposite Side
adjacent side of A / Opposite side of A
hypotenuse
Adjacent Side
So Tangent ( Tan ) and Cotangent ( Cot ) are called multiplicative inverse trigonometrical ratioes of each other .

18. Ratioes of trigonometrical Ratioes.
Opposite side
hypotenuse
Adjacent Side
= Opposite side of A / adjacent side of A

19. Ratioes of trigonometrical Ratioes.
hypotenuse
Opposite Side
Adjacent Side
adjacent side of A / Opposite side of A

20. Think – Discuss equal to Tan A ? Is
Opposite Side
hypotenuse
Adjacent Side
= Opposite side of A / adjacent side of A

21. Think – Discuss equal to Cot A ? Is
Opposite Side
hypotenuse
Adjacent Side
adjacent side of A / Opposite side of A

22. adjacent side of A = AB = 4k
Example: 1 If
then find the other trigonometric ratio of angle A.
= Opposite side of A / adjacent side of A C
Opposite side : adjacent side = 3:4
Opposite Side 3k 5k
hypotenuse
Opposite Side of A = BC = 3k
( where k is any positive number )
adjacent side of A = AB = 4k A
Adjacent Side B 4k
According to Pythagoras theorem in Δ ABC AC2 = AB2+BC2
AC2 = (3k)2+(4k)2 = 9k2+16k2 = 25k2 = (5k)2
AC = 5k

23. Example-2
If A and P are acute angles such that Sin A = Sin P then prove that A = P . R C
If Sin A = Sin P A P Q B
Let
and
then If

24. Another Method of Example-2
If A and P are acute angles such that Sin A = Sin P then prove that A = P .
In ΔACP , Given that
Sin A = Sin P þ P
For angle A in ΔACP
For angle P in ΔACP A C

25. Consider a triangle PQR , right angled at P in which
Example : 3
Consider a triangle PQR , right angled at P in which
PQ = 29 units , QR= 21 units and PQR =  then
find the value of Q
29 units
and
21 units
According to Pythagoras theorem in Δ PQR PR2 = PQ2 - QR2 R P
20 units
PR = 20 units

26. Exercise – 11.1 In a right angle triangle ABC, 8 cm , 15 cm and
17 cm are the lengths of AB,BC and CA
respectively. Then find out Sin A , Cos A
and tan A . C
17 cm
15cm
Given that , AB = 8 cm ; BC = 15 cm ;
CA = 17 Cm in a right angle triangle ABC . B A
8 cm
CA is the longest side of a right angle triangle ABC . Hence CA is the hypotenuse of a ΔABC.
For angle A
Sin A = Opposite side to A/ hypotenuse =
Cos A = Adjacent side to A/ hypotenuse =
tan A = Opposite side to A / Adjacent side to A =

27. Solution : Given that in ΔPQR , Q = 900 , PQ = 7cm , PR = 25 cm .
Exercise – 11.1
2.The sides of a right angle triangle PQR are PQ = 7 cm , QR = 25 cm and Q = 900 respectively. Then Find tan Q – tan R . R
Solution : Given that in ΔPQR , Q = 900 , PQ = 7cm , PR = 25 cm .
24cm
25 cm Q P
7 cm
According to Pythagoras theorem in ΔPQR
QR2 = PR2 - PQ2 = = = 576=242
QR = 24 cm
For angle P ; tan P = Opposite side to P / Adjacent side to P =
For angle R ; tan R = Opposite side to P / Adjacent side to P =

28. Cos  = Adjacent side to / hypotenuse =
Exercise – 11.1
3. In a right angle triangle ABC with right angle at B, in which a = 24 units , b = 25 units and BAC = . Then find cos  and tan  . C
b = 25 units
a = 24 units  B A
7 units
Solution : Given that , in ΔABC B = 900 a = BC= 24 units , b = AC = 25 units and BAC =  .
According to Pythagoras theorem in ΔABC AB2 = AC2 - BC2 = = = 49=72
AB = 7 units
Cos  = Adjacent side to / hypotenuse =
tan  = Opposite side to / Adjacent side to  =

29. Solution : Let in triangle ΔABC ,B = 900 Cos A = 12/13
Exercise – 11.1 C
13k
Solution : Let in triangle ΔABC ,B = 900 Cos A = 12/13 5k B A
12k
Cos A = 12/13 = Adjacent side to A/ hypotenuse
Adjacent side : hypotenuse = 12:13
Adjacent side to A = AB =12k ( where k is any positive number )
hypotenuse = AC =13k
According to Pythagoras theorem in ΔABC
BC2 = AC2 - AB2 = (13k)2 – (12k)2 =169k2 -144k2 = 25k2 =(5k)2
BC = 5k
Sin A = Opposite side to A / hypotenuse =
tan A = Opposite side to A / adjacent side to A =

30. 5. If 3 tan A = 4 then find Sin A and Cos A . C
Exercise – 11.1
5. If 3 tan A = 4 then find Sin A and Cos A . C
Solution : Given that , 3 tan A = 4 5k
tan A = 4 / 3 = Opposite side to A / adjacent side to A 4k
Opposite side : adjacent side = 4:3 B A 3k
Opposite side to A = BC =4k ( where k be any positive number )
Adjacent side to A = AB =3k
According to Pythagoras theorem in ΔABC
AC2 = AB2 + BC2 = (3k)2 + (4k)2 =9k k2 = 25k2 =(5k)2
AC = 5k
Sin A = Opposite side to A / hypotenuse =
Cos A = adjacent side to A / hypotenuse =

31. Exercise – 11.1
6. If A and X are acute angles such that Cos A = Cos X then show that A = X .
Given that , in triangle ΔACX Cos A = Cos X. X
For angle A in ΔACX A C
For angle X in ΔACX

32. = Adjacent side to  / Opposite side to  B A
Exercise – 11.1 C 8k
Given that Cot  = 7 / 8
= Adjacent side to  / Opposite side to   B A 7k
Adjacent side : Opposite side = 7:8
Adjacent side to  = AB = 7k ( Where k be any positive number)
Opposite side to  = BC =8k
According to Pythagoras theorem in ΔABC
AC2 = AB2 + BC2 = (7k)2 + (8k)2 = 49k k2 = 113k2 =
Sin = Opposite side to / hypotenuse =
Cos  = Adjacent side to / hypotenuse =

34. 8. In a right angle triangle ABC , right angle at B , if C
Exercise – 11.1
8. In a right angle triangle ABC , right angle at B , if C
then find the values of
i) Sin A Cos C+ Cos C Sin A
ii) Cos A Cos C – Sin A Sin C
Solution : Given that , in a right angle triangle ABC , right angle at B and B A k
= Opposite side to A / hypotenuse
Opposite side : adjacent side =
Opposite side to A = BC =
( where k be any positive number )
Adjacent side to A = AB =1k = k
According to Pythagoras theorem in ΔABC AC2 = AB2 + BC2 =
Sin A= Opposite side to A / hypotenuse =
Sin C = Opposite side to C / hypotenuse =

35. Cos A= Adjacent side to A / hypotenuse = C
Cos C= Adjacent side to C / hypotenuse =
i) Sin A Cos C+ Cos C Sin A B A k
ii) Cos A Cos C – Sin A Sin C

36. 11.3 Trigonometric Ratioes of Some Specific Angles
Trigonometric Ratio of 450
In isosceles right angle triangle ABC , right angle at B A=C=450 and Let BC=AB = a
By Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2 = B A
Sin 450 = Opposite side to 450 / hypotenuse =
Cos 450 = Adjacent side to 450 / hypotenuse =
Tan 450 = Opposite side to 450 / Adjacent side to 450 =
Cot 450 = Adjacent side to 450 / Opposite side to 450 =
Sec 450 = hypotenuse / Adjacent side to 450 =
Cosec 450 = hypotenuse / Opposite side to 450 =

37. 11.3.2 Trigonometric Ratioes of 300 and 600
Consider an equilateral triangle ABC. In Δ ABC A=B=C= 600 and Let AB=BC=CA = 2a A
Draw a perpendicular line AD from vertex A to BC as shown in adjacent figure . Perpendicular AD acts as angle bisector of angle A and bisector of the the side BC in the equilateral triangle ABC . D B C
Therefore , ΔABC BAD =CAD=300
and point D divides the side BC into equal halves. BD= ½ BC = ½ . 2a = a
By Pythagoras theorem in Δ ABD AD2 = AB2 - BD2 =

38. Sin 300 = Opposite side to 300 /hypotenuseA
Cos 300 = Adjacent side to 300 /hypotenuse
Tan 300 = Opposite side to300 /Adjacent side to 300 D B
Cot 300 = Adjacent side to 300/ Opposite side to 300
Sec 300 = hypotenuse / Adjacent side to 300
Cosec 300 = hypotenuse / opposite side to 300

39. Sin 600 = Opposite side to 600 / hypotenuseA
Cos 600 = Adjacent side to 600 / hypotenuse
Tan 600 = Opposite side to 600/ Adjacent side to 600 D B
Cot 600 = Adjacent side to 600 / Opposite side to 600
Sec 600 = hypotenuse / Adjacent side to 600
Cosec 600 = hypotenuse / opposite side to 600

40. 11.3.3 Trigonometric Ratio of 00
Suppose a Segment AC of length r is making an acute angle with ray AB. Height of C from B is BC. C
When AC leans more on AB so that the angle made by it decreases . As the angle A decreases , the height of C from AB ray decreases and foot B is shifted from B to B1 and B2 gradually when the angle becomes zero, height ( i.e.opposite side of the angle ) will also become zero and adjacent side would be equal to r. A B C C C B B1 A B2 A
If A = 00 then BC = 0 and AC = AB = r

41. 11.3.3 Trigonometric Ratio of 900 C C C
When angle made by AC with ray AB increased, height of Point C increases and the foot of the perpendicular shifts from B to Y and then to X and so on. C
Height BC increases gradually , the angle on C gets continuous increment and at one stage the angle reaches 900. At that time , point B reaches A and AC equal to BC . A Y B X C C C
When the angle becomes 900 , base ( i.e. adjacent side of the angle ) would become zero , height of C from AB ray increases and it would be equal to AC and that is the length equal to r . B A B A Y
If A = 900 then AB = 0 and AC = BC = r

42. Do this
Find Cosec 600 , Sec 600 and Cot 600 .

43. Think – Discuss 1. What can you say about ? Is it defind ? Why?
Reason : Division by zero is not Possible So it is not defined .
2. What can you say about
? Is it defind ? Why?
Reason : Division by zero is not Possible So it is not defined .
? Why?

44. Table of Trigonometric Ratioes00
300
450
600
900
Particulars
Step : 1 1 2 3 4
Write from 0 to 4 serially
Step : 2
Divide each one by 4
Step : 3
Find the square root of each
Step : 4
Simplify
Sin A
Sin Value
Cos A
Write Sin values reversely
Tan A
Cot A
Write Tan Values reversly
Sec A
Cosec A
Write Sec Values reversly

45. Think – Discuss
What can you say about the value of Sin A and Cos A , as the value of angle A increases from 00 to 900? ( Observe the above table )
The Following table gives the evidence to say the given statement is true. If angle A increases then its sine values are also increases. A 00
300
450
600
900
Sin A
Given statement is False. Because if value of angle A increases then its cos values are decreases. A 00
600
900
300
450
Cos A

46. Example - 4
In ΔABC, right angle is at B , AB= 5 cm and ACB=300 , Determine the lengths of the sides BC and AC.
Solution: Given that in ΔABC, right angle at B , AB = 5 cm and ACB=300 A
Opposite side to 300 = AB = 5 cm
Adjacent side to 300 = BC
5 cm B C cm
By using Pythagoras theorem in Δ ABC , AC2 = AB2 + BC2 cm

47. Example- 5
A Chord of a circle of radius 6 cm is making an angle 600 at the centre. Find the length of the chord?
Solution : Given that radius of the circle OA=OB=6 cm
AOB = 600
OC is height from O upon AB and it is angle bisector
6 òÜ….Ò$.
6 òÜ….Ò$.
COB=COA= 300
Length of the chord AB = 2AC=2BC A B C
Length of the chord AB = 2BC=2(3)=6 cm

48. Example:6
In Δ PRQ, right angle is at Q , PQ = 3cm and PR = 6 cm. Determine QPR and PRQ .
Solution : Given PQ = 3 cm and PR = 6cm P
For the angle of R in ΔPQR
6 cm
3 cm Q R

49. Example : 7
A > B , Find A and B .

50. Exercise 11.2
1. Evaluate the following

51. Exercise 11.2

52. c Exercise 11.2 2.Choose the right option and justify your choice. ( )
( ) c
(a) Sin 600
(b) Cos 600
(c) tan 300
(d) Sin 300

53. d c Exercise 11.2 2.Choose the right option and justify your choice.
( ) d
(c) Sin 450
(d) 0
(b) 1
(a) tan 900
( ) c
(d) Sin 300
(a) Cos 600
(b) Sin 600
(c) tan 600

54. Exercise 11.2
3.Evaluate Sin 600cos cos 600 Sin What is the value of
Sin ( ). What can you conclude ?
Solution : Sin 600cos Cos 600 Sin 300
………….1
Sin ( ) = Sin 900 =1 ………….2
From 1 and 2
Sin ( ) =
Sin 600cos Cos 600 Sin 300
Sin ( A + B ) =
Sin A cos B + Cos A Sin 300

55. Exercise 11.2
4. Is it right to say Cos( ) = cos 600 Cos 300 – Sin 600 Sin 300
Solution : Cos600 cos 300 – Sin 600 Sin 300
………….1
Cos ( ) = Cos 900 = 0 ………….2
From 1 and 2
Cos 600cos 300 – Sin 600 Sin 300
Cos ( ) =
Therefore it is right to say
Cos( ) = cos 600 Cos 300 – Sin 600 Sin 300

56. Exercise 11.2
5. In a right angle triangle ΔPQR, right angle is at Q and PQ = 6 cm , RPQ = Determine the lengths of QR and PR .
Solution : Given that right angle at Q in ΔPQR and PQ = 6 cm , RPQ = 600 P
For angle P in ΔPQR
12 cm
6 cm Q R cm cm cm

57. Exercise 11.2
6. In ΔXYZ, right angle is at Y , YZ = x and XY = 2x then determine YXZ and YZX
Solution : In ΔXYZ , right angle is at Y , XZ = 2x , XY = x X 2x
By using Pythagoras theorem XZ2 = XY2 + YZ2
(2x)2 = (x)2 + YZ2 x
4x2 = x2 + YZ2
YZ2 = 4x2 – x2 Y Z
YZ2 = 3x2
For angle Z in ΔXYZ
For angle X in ΔXYZ

58. Exercise 11.2
7. Is it right to say that Sin (A+B) = Sin A + Sin B ? Justify your answer
Solution : Let A = 600 and B = 300
LHS Sin (A+B) = Sin ( ) = Sin 900 = 1
RHS Sin A + Sin B = Sin Sin 300 =
It is not right to say that Sin (A+B) = Sin A + Sin B

59. For which value of Think – Discuss For which value of acute angle
Is true ?
For which value of
, above equation is not defined?
Solution :
For = 600 the given statement is true .

60. Trigonometric Ratioes of complementary Angles
Two angles are said to be complementary , if their sum is equal to 900 C
900 - x
hypotenuse
Adjacent side to x
Opposite side to x x A B
Adjacent side to x
Opposite side to x
Let

62. Example: 8
Evaluate
Solution:

63. Example : 9 Where 7A is an acute angle, Find the value of A .
Solution : Given
Since 7A is an actute angle, ( 900 – 7A ) and ( A – 60 ) are also actute

64. Given that Sin A = Cos B Example : 10 Example : 11
If Sin A = Cos B , then Prove that A + B = 900.
Given that Sin A = Cos B
Solution :
Express Sin Tan 810 in terms of trigonometric ratioes of angle between 00 and 450
Example : 11

65. Example : 11
If A,B and C are interior angles of right angle triangle ABC then Show that
Solution :
Given A,B and C are interior angles of right triangle ABC then
On taking Sin ration on both sides

66. Exercise 11.3
1. Evaluate
Solution :

67. Exercise 11.3
2. Show that
Solution:

68. Exercise 11.3 Another Method of Solution Solution:
3. If Tan 2A = Cot (A – 180), where 2 A is an acute anlgle. Find the value of A .
Solution :
Tan 2A = Cot (A – 180)
Cot ( 900 – 2A) = Cot (A – 180)
Cot A = Cot B AÆÿ$$¯èþ A=B
( 900 – 2A) = (A – 180)
= A + 2A
3A = 1080

69. Exercise 11.3
4. If Tan A = Cot B where A and B are actute angles , Prove that A+B=900
Solution :
Given that Tan A = Cot B

70. Exercise 11.3
5. If A,B and C are interior angles of a triangle ABC, then show that
Solution :
Given that A,B and C are interior angles of a triangle ABC
By taking tan ratio on both sides

71. Exercise 11.3
6. Express Sin cos 650 in terms trigonometric ratioes of angles between 00 and 450

72. 11.5 Trigonometric Identities
Consider a right angle triangle ABC with right angle at B . From Pythagoras theorem AC2 + AB2 = AC2 C
Dividing each term by AC2 B A
required trigonometric identity .

73. 11.5 Trigonometric Identities
Consider a right angle triangle ABC with right angle at B . From Pythagoras theorem AC2 + AB2 = AC2 C
Dividing each term by AB2 B A
required trigonometric identity

74. 11.5 Trigonometric Identities
Consider a right angle triangle ABC with right angle at B . From Pythagoras theorem AC2 + AB2 = AC2 C
Dividing each term by BC2 B A
required trigonometric identity

76. Do this
then finc Cos A
then find Sec x
then find cot 

77. Try This
Evaluate the following and Justify your answer.

78. Example : 13
Show that Cot  + tan  = Sec  Cosec 
Example : 14
Show that tan2  + tan4  = Sec4  – sec2 

79. Example : 15
Show that

80. Exercise :11.4
1. Evaluate the following
Solution:

81. Exercise :11.4
1. Evaluate the following
Solution:

82. Exercise : 11.4

83. Exercise : 11.4

84. Exercise : 11.4
Another Method

85. Exercise : 11.4

86. Exercise : 11.4

87. Exercise : 11.4
Given that

88. Exercise : 11.4
Dividing by
Adding
and
Subtracting
from

89. Exercise : 11.4
Another method
Given that

90. Optional Exercise

91. Optional Exercise
Dividing Numeratior and Denominator by Cos

92. Optional Exercise

93. Optional Exercise

94. Optional Exercise

95. Optional Exercise