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( 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.
1
Relations between sides and angles of a right triangle
An angle which contains 900 then
that angle is said to be right angle .
Opposite side
If an angle of a triangle contains
900 , then that triangle said to be
right angle triangle.
 Adjecent Side
90
0
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 .
2
Ratioes of a right triangle.
BC
a
Sin A 
  Opposite side to A / hypotenuse
AB
c
Adjacent side
Opposite side
The ratio of opposite side and hypotenuse is called Sine A .
Shortly it is written as Sin A and read as Sine A.
900
The ratio of adjacent side and hypotenuse is called Cosine A .
Shortly it is written as Cos A and read as Cos A
AC
b
Cos A 
  adjacent side to A / hypotenuse
AB
c
3
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
BC a
Tan A 

AC b
= Opposite side to A / adjacent side to A
Adjacent side
90 0
The ratio of adjacent side and Opposite side is called Cotangent A .
Shortly it is written as Cot A and read as Cot A
AC b
Cot A 
  adjacent side to A / Opposite side to A
BC a
4
AB C
Sec A 
 = hypotenuse / adjacent side to A
AC b
Opposite Side
The ratio of hypotenuse and adjacent side is called Secant A .
Shortly it is written as Sec A and read as Sec A
Adjacent Side 90 0
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
AB C
Cosec A 

BC a
= hypotenuse / Opposite side to A
5
1. Identify hypotenuse”, “Opposite
Side” and “adjacent side” for angle R
in the given triangle.
for angle R in the adjacent triangle.
PR = hypotenuse
Opposite Sice
Do this
P
Q
R
Adjacent Side
PQ = Opposite Side
Z
QR = adjacent Side
2 (i) Identify hypotenuse”, “Opposite
Side” and “adjacent side” for angle X
in the given triangle.
Y
hypotenuse
for angle X in the adjacent triangle.
XY = hypotenuse
YZ = Opposite side
XZ =adjacent side
6
X
Do this
2 (i) Identify hypotenuse”, “Opposite
Side” and “adjacent side” for angle Y
in the given triangle.
for angle Y in the adjacent triangle.
Z
Y
X
hypotenuse
XY = hypotenuse
XZ = Opposite Side
YZ = adjecent Side
7
{ç³Äæý$†²…^èþ…y
Write
lengths of “hypotenuse”, “Opposite
Side”
ìþ: and “adjacent side” for the given
angles in the given triangle.
4 cm
1. For angle C 2. For angle A
According Pythagoras Theorem
AC 2  AB 2  BC 2
B
AB  AC  BC  5  4  25  16  9  3
2
2
2
2
Adjacent Side
C
2
2
AB  3
5 cm
A
Opposite side
3 cm
( i ) For angle C
AC = hypotenuse = 5 cm
AB = Opposite Side = 3 cm
BC = adjacent Side = 4 cm
( ii ) For angle A
4 cm
AC = hypotenuse = 5 cm
BC = Opposite Side = 4 cm
AB =adjacent side = 3 cm
Opposite Side
C
B
5 cm
Adjacent Side
3 cm
A
8
C
Do This
1. Find (i) Sin C (ii) Cos C and
(iii) tan C in the adjacent triangle.
In the adjacent Δ ABC , B = 900
A
B
12 cm
According Pythagoras Theorem
 AC  AB  BC
2
2
2
AB  AC  BC
2
2
2
AB  13  5  169  25  144  12
2
2
2
 AB  12 cm
AB 12
(i) Sin C 

AC 13
AB 12
(iii ) Tan C 

BC 5
2
BC
5
(ii ) Cos C 

AC 13
9
X
Do this
In triangle XYZ , Y is right angle.
XZ= 17cm and YZ=15 cm then find
(i) sin X (ii) Cos Z (iii ) tan X .
In triangle XYZ , Y is right angle.
According Pythagoras Theorem
 XZ 2  XY 2  YZ 2
XY  XZ  YZ
2
2
Y
2
XY  17  15  289  225  64  8
2
2
Z
15 cm
2
2
XY = 8 cm
YZ 15
(i ) Sin X =

XZ 17
YZ 15
(ii ) Cos Z =

XZ 17
YZ 15
(iii ) Tan X=

XY 8
10
P
Do This
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 .
x
In triangle PQR with right angle at Q
According Pythagoras Theorem
Q
 PR 2  PQ 2  QR 2
 PR  7  24  49  576  625  25
2
2
2
R
24 cm
2
PR = 25 cm
QR 24
(i ) Sin x =

PR 25
PQ
7
(ii ) Cos x =

PR 25
11
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 .
8 cm
A
Try This
In a right angle triangle ABC ,
right angle is at C .
C
BC  CA  BC  CA  23  7  30
2BC  30
15
30
BC 
 15
2
B
15 cm
In ΔABC , C = 900
According Pythagoras Theorem AB 2  AC 2  BC 2
BC  CA  23
15  CA  23
CA  23 15  8
AB  8  15  64  225  289  17
2
2
2
AB = 17 cm
BC 15
Sin A 

AB 17
AC
8
Tan B 

BC 15
12
2
Think - Discuss
4
deos exists for some value of angle x ?
3
Sin value exists always less than 1.
4
Sin x  Value is more than 1. So it does not exists for some
3
value of angle x .
(i) Sin 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 .
AP b
Sin 
 b
OP 1
Y
P ( a, b)
1
Y Coordinate
OA a
Cos 
  a  X Coordinate
OP 1
X
o
(0, 0)

aA
b
X
Y
13
P is making an angle 3600 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 C ( 1, 0)
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 .
B (0,1)
o
A(1, 0)
D(0, 1)
According to the Ordered Pairs
Sin 00 = 0
Sin 900 = 1
Sin 1800 = 0
Cos 00 = 1
Cos 900 = 0
Cos 1800 = – 1
Sin 2700 = – 1
Cos 2700 = 0
Sin 3600 = 0
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 .
14
Multiplicative Inverses of Trigonometrical Ratioes
BC a
Sin A 
 
AB c
AB C
Cosec A 

BC a
Opposite side of A / hypotenuse
= hypotenuse / Opposite side of A
Adjacent Side
So Sine and Cosec are called multiplicative inverse trigonometrical
ratioes of each other .
1
Co sec A 
Sin A
1
Sin A 
Co sec A
15
Multiplicative Inverses of Trigonometrical Ratioes
AC b
Cos A 
 
AB c
AB C
Sec A 

AC b
Adjacent side of A / hypotenuse
= hypotenuse / Adjacent side of A
Adjacent Side
So Cosine (Cos) and Secant (Sec) are called multiplicative inverse
trigonometrical ratioes of each other .
1
1
Sec A 
Cos A 
Sec A
Cos A
16
Multiplicative Inverses of Trigonometrical Ratioes
= Opposite side of A / adjacent side of A
Opposite Side
BC a
Tan A 

AC b
AC b
Cot A 
 
BC a
adjacent side of A / Opposite side of A
Adjacent Side
So Tangent ( Tan ) and Cotangent ( Cot ) are called multiplicative
inverse trigonometrical ratioes of each other .
1
Cot A 
Tan A
1
Tan A 
Cot A
17
Ratioes of trigonometrical Ratioes.
BC
Sin A
BC
AB



AC
Cos A
AC
AB
Adjacent Side
= Opposite side of A / adjacent side of A
 Tan A
Sin A
 Tan A
Cos A
18
Ratioes of trigonometrical Ratioes.
AC
Cos A
AC
AB



BC
Sin A
BC
AB
Adjacent Side
adjacent side of A / Opposite side of A
 Cot A
Cos A
 Cot A
Sin A
19
Think – Discuss
Sin A
equal to Tan A ?
Is
Cos A
BC
Sin A
BC
AB



Cos A AC
AC
AB
Adjacent Side
= Opposite side of A / adjacent side of A
 Tan A
Sin A
 Tan A
Cos A
20
Think – Discuss
Is
Cos A
equal
Sin A
to Cot A ?
AC
Cos A
AC
AB



BC
Sin A
BC
AB
Adjacent Side
adjacent side of A / Opposite side of A
 Cot A
Cos A
 Cot A
Sin A
21
Example: 1
If
Tan A 
3
4
then find the other trigonometric ratio of angle A.
3
Tan A  = Opposite side of A / adjacent side of A
4
C
Opposite side : adjacent side = 3:4
Opposite Side of A = BC = 3k
( where k is any positive number )
adjacent side of A = AB = 4k
A
3k
Adjacent Side
4k
B
According to Pythagoras theorem in Δ ABC AC2 = AB2+BC2
AC2 = (3k)2+(4k)2 = 9k2+16k2 = 25k2 = (5k)2
AC = 5k
AC 5k 5
BC 3k 3
Cosec A 


SinA 


BC 3k 3
AC 5k 5
AB 4k 4
Cos A 


AC 5k 5
BC 3k 3
Tan A 


AB 4k 4
AC 5k 5
Sec A 


AB 4k 4
AB 4k 4
Cot A 


BC 3k 3
22
Example-2
If A and P are acute angles such that Sin A = Sin P
then prove that A = P .
BC
R
C
Sin A 
AC
QR
Sin P 
PR
A
BC QR
Let

k
AC PR
BC QR


AC
PR
BC
 k  BC  k . AC
AC
AB


PQ
If
AC 2  BC 2
PR  QR
2
Q
P
B
If Sin A = Sin P
2

QR
 k  QR  k .PR
PR
and
AC 2  k 2 AC 2
PR  k PR
2
AC
AB
BC



PR
PQ
QR
2
2
then
A  P

AC 2 (1  k 2 )
PR (1  k )
2
ABC
2

AC 2
AC

2
PR
PR
PQR
23
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
P
A
C
Sin A = Sin P þ
CP
For angle A in ΔACP Sin A 
AP
AC
For angle P in ΔACP Sin P 
AP
CP
AC
Sin A = Sin P 

AP
AP
 CP  AC
A  P
24
Q

20 units
21 units
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
2
2
2
2
(
ii
)
Cos


Sin

and
(i )Cos   Sin 
According to Pythagoras theorem in
P
Δ PQR PR2 = PQ2 - QR2
R
PR 2  292  212  841  441  400  202
PR = 20 units
PR
20
Sin 

PQ
29
QR
21
Cos 

PQ
29
2
2
21
20
441 400 841
2
2
(i)Cos   Sin   2  2 


1
29
29
841 841 841
2
2
21
20
441 400 41
2
2
(ii)Cos   Sin   2  2 


29 29 841 841 841
25
Exercise – 11.1
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 .
Given that , AB = 8 cm ; BC = 15 cm ;
CA = 17 Cm in a right angle triangle ABC .
A
15cm
C
B
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 =
BC
15

AC
17
AB
8

AC
17
tan A = Opposite side to A / Adjacent side to A = BC  15
AB
8
26
Exercise – 11.1
2.The sides of a right angle triangle PQR are PQ = 7 cm , QR = 25
R
cm and Q = 900 respectively. Then Find tan Q – tan R .
24cm
Solution : Given that in ΔPQR , Q = 900 ,
PQ = 7cm , PR = 25 cm .
P
7 cm
Q
According to Pythagoras theorem in ΔPQR
QR2 = PR2 - PQ2 = 252 - 72 =625-49 = 576=242
QR = 24 cm
For angle P ; tan P = Opposite side to P / Adjacent side to P =
QR
24

PQ
7
For angle R ; tan R = Opposite side to P / Adjacent side to P =
PQ
7

QR
24
24 7
242  7 2 576  49 527
tan P  tan R 




7 24
7.24
168
168
27
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  .
a = 24 units
C

A
7 units
B
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 = 252 - 242 =625-576 = 49=72
AB = 7 units
Cos  = Adjacent side to / hypotenuse = AB  7
AC
25
tan  = Opposite side to / Adjacent side to  = BC  24
AB
7
28
Exercise – 11.1
C
5k
12
4. If cos A 
, Find Sin A and tan A .
13
Solution : Let in triangle ΔABC ,B = 900
Cos A = 12/13
A
Cos A = 12/13 = Adjacent side to A/ hypotenuse
12k
B
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 =
BC
5k
5


AC 13 k 13
tan A = Opposite side to A / adjacent side to A =
BC 5 k
5


AB 12 k 12
29
Exercise – 11.1
5. If 3 tan A = 4 then find Sin A and Cos A .
C
4k
Solution : Given that , 3 tan A = 4
tan A = 4 / 3 = Opposite side to A / adjacent side to A
Opposite side : adjacent side = 4:3
B
3k
A
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 =9k2 + 16k2 = 25k2 =(5k)2
AC = 5k
BC
4k
4


Sin A = Opposite side to A / hypotenuse = AC 5 k
5
Cos A = adjacent side to A / hypotenuse =
AB 3 k
3


AC 5 k
5
30
Exercise – 11.1
6. If A and X are acute angles such that Cos A = Cos X then
show that A = X .
X Given that , in triangle ΔACX Cos A = Cos X.
For angle A in ΔACX
A
AC
Cos A 
AX
C
For angle X in ΔACX
CX
Cos X 
AX
AC
CX
Cos A = Cos X 

AX
AX
 AC  CX
 CX  AC
A   X
31
Exercise – 11.1
7
7. Given Cot  , then evaluate
8
(1  Sin  )(1  Sin  )
(1  Sin  )
i.
ii.
(1  Cos  )(1  Cos  )
Cos 
C
Given that Cot  = 7 / 8
= Adjacent side to  / Opposite side to  A
Adjacent side : Opposite side = 7:8
8k
113k

7k
B
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 = 49k2 + 64k2 = 113k2 = ( 113k )2 
AC  113k
Sin = Opposite side to / hypotenuse =
BC
8k
8


AC
113 k
113
Cos  = Adjacent side to / hypotenuse = AB  7 k
AC
7

113 k
113
32
113k
(1  Sin  )(1  Sin  )
1  Sin 
i.

2
(1  Cos  )(1  Cos  ) 1  Cos 
2
2
 8 
64
113  64
1 
1

49
113 

113
113




2
49
113  49
64
 7 
1

1 

113
113
113


8
1
(1  Sin  )
113 
ii.

7
Cos 
113
113  8
113
7
113
8  113

7
33
Exercise – 11.1
8. In a right angle triangle ABC , right angle
at B , if tan A  3
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
A
triangle ABC , right angle at B and
C
2k
3k
k
B
3
= Opposite side to A / hypotenuse
1
Opposite side : adjacent side = 3 :1
tan A  3 
Opposite side to A = BC = 3k ( where k be any positive number )
Adjacent side to A = AB =1k = k
According to Pythagoras theorem in ΔABC AC2 = AB2 + BC2 =

3k

2
  k   3k 2  k 2  4k 2   2k 
2
2
AC  2k
Sin A= Opposite side to A / hypotenuse =
Sin C = Opposite side to C / hypotenuse =
3k
3

2k
2
k
1

2k
2
34
Cos A= Adjacent side to A / hypotenuse =
k
1

2k
2
C
Cos C= Adjacent side to C / hypotenuse =
2k
3k
3k
3

2k
2
i) Sin A Cos C+ Cos C Sin A
A
k
B
1
3 3 1 1 3 1 3 1 4


  
 1
2 2 22 4 4
4
4
ii) Cos A Cos C – Sin A Sin C
1 3
31
3
3




0
2 2
2 2 4
4
35
11.3 Trigonometric Ratioes of Some Specific Angles
C
11.3.1 Trigonometric Ratio of 450
In isosceles right angle triangle ABC , right angle at B
A=C=450 and Let BC=AB = a
2a
By Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2 = a  a  2a 
2
2
2
 2a 
450
a
2
a
A
AC  2a
B
1
Sin
450 =
Opposite side to
450 /
hypotenuse =
AB
a
1


AC
21a
2
BC
a
1


Cos
side to
/ hypotenuse AC 2a 2

1
=
AB
a

0
0
0
 1
Tan 45 = Opposite side to 45 / Adjacent side to 45 =
BC
a

1
BC
a
0
0
0
 1
Cot 45 = Adjacent side to 45 / Opposite side to 45 =
AB a
450 = Adjacent
450
AC
2a

 2
Sec
hypotenuse / Adjacent side to
=
AB
a
Cosec 450 = hypotenuse / Opposite side to 450 = AC  2a  236
BC
a
450 =
450
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
2a
adjacent figure . Perpendicular
AD acts as angle bisector of
angle A and bisector of the the
0
60
side BC in the equilateral triangle
ABC .
B
0
30
30
0
2a
3a
600
a
Therefore , ΔABC BAD =CAD=300
and point D divides the side BC into equal
halves. BD= ½ BC = ½ . 2a = a
D
2a
a
C
By Pythagoras theorem in Δ ABD AD2 = AB2 - BD2 =
  2a    a   4a  a  3a 
2
2
2
2
2
 3a 
2
AD  3a
37
1
Sin 300 = Opposite side to 300 /hypotenuse

Cos 300 = Adjacent side to 300 /hypotenuse 
BD
a
1



AB
2a
2

AD
3a
3


AB
2a
2
Tan 300 = Opposite side to300 /Adjacent side to 300
1
300
2a
BD
a
1



AD
3a
3
3a
a
B
Cot 300 = Adjacent side to 300/ Opposite side to 300
AD
3a
3



 3
BD
a
1
Sec
300 =
hypotenuse / Adjacent side to
A
300
D
AB 2a 2


 2
BD
a
1
1
Cosec 300 = hypotenuse / opposite side to 300 
AB
2a
2



AD
3a
3

38
Sin 600 = Opposite side to 600 / hypotenuse  AD  3a  3
AB
2a
2
A
0
0
Cos 60 = Adjacent side to 60 / hypotenuse
1
BD
a
1



AB 2a 2
Tan 600 = Opposite side to 600/ Adjacent side to 600
AD
3a
3



 3
BD
a
1
2a
3a
600
a
B
1
D
Cot 600 = Adjacent side to 600 / Opposite side to
1
600

Sec
600 =
BD

AD
a
1


3a
3

hypotenuse / Adjacent side to
600
AB 2a 2


 2
BD a
1
1
Cosec
600 =
hypotenuse / opposite side to
600

AB
2a
2


AD
3a
3
39
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
r
from B is BC.
When AC leans more on AB

so that the angle made by it
A
decreases . As the angle A
r
decreases , the height of C
C 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 )
B B1 A
A
will also become zero and
adjacent side would be equal
to r.
C
0
B
C
B2
If A = 00 then BC = 0 and AC = AB = r
BC
0

0
AC
r
AB
r
Cos 00 
 1
AC
r
Sin 00 
BC 0
Tan 0 
 0
AB r
0
Cosec 00 
AC r
 
BC 0
Sec 00 
AC
r
 1
AB
r
AB
r
Cot 00 


BC
0
40
11.3.3 Trigonometric Ratio of 900
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.
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 .
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
A
equal to r .
A
C
X
C
C
C
Y
B
C
B A
Y
B
If A = 900 then AB = 0 and AC = BC = r
BC r
Sin 90 
 1
AC r
AB 0
0
Cos 90 
 0
AC r
0
BC r
Tan 90 
 
AB 0
0
AC r
Cosec 90 
 1
BC r
0
AC r
Sec 90 
 
AB 0
AB 0
0
Cot 90 
 0
BC r
0
41
Do this
Find Cosec 600 , Sec 600 and Cot 600 .
1
1
2
2
Co sec 60 

 1.

0
sin 60
3
3
3
2
0
1
1
2
2
Se c 30 

 1.

0
Cos 30
3
3
3
2
1
0
cot 60
0
Cos 600


0
Sin 60
 
2
3

2
1
3
42
Think – Discuss
1
1. What can you say about Co sec 0  Sin 00 ? Is it defind ? Why?
1
1
Co sec 00 


0
Sin 0 0
0
Reason : Division by zero is not Possible So it is not defined .
1
2. What can you say about Cot 0 
? Is it defind ? Why?
0
Tan 0
0
1
1
Cot 0 
 
0
Tan 0
0
0
Reason : Division by zero is not Possible So it is not defined .
3. Sec 00  1? Why?
1
1
Sec 0 
 1
0
Cos 0
1
0
43
Table of Trigonometric Ratioes
A
00
300 450
600 900
Step : 1
0
1
2
3
4
Step : 2
0
4
1
4
2
4
3
4
4
4
Divide each one by 4
Step : 3
0
4
1
4
2
4
3
4
4
4
Find the square root of each
Step : 4
0
0
4
1 1

4 2
Sin A
0
1
2
1
2
3 3

4 2
3
2
Cos A
1
3
2
1
1
2
1
2
1
Tan A
0
Cot A
Sec A

1
Cosec A

3
3
2
3
2
2 1

4
2
2
4
1
4
1
0

1
3
1
1
3
0
2
2

2
2
3
1
Particulars
Write from 0 to 4 serially
Simplify
Sin Value
Write Sin values reversely
Sin
Cos
Write Tan Values reversly
1
Cos
Write Sec Values reversly
44
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 )
(i) If A  B then SinA  Sin B. Is it true ?
Solution : Given statement is If A  B then SinA  Sin B.
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
Sin A
00
300
450
0
1
2
1
2
600 900
3
2
1
(ii) If A  B then Cos A  Cos B. Is it true ?
Solution : Given statement is If A  B then Cos A  Cos B.
Given statement is False. Because if value of angle A increases
then its cos values are decreases.
A
00 300 450 600 900
Cos A
1
3
2
1
2
1
2
0
45
Example - 4
Adjacent side to 300 = BC
5 cm
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
A
B , AB = 5 cm and ACB=300
Opposite side to 300 = AB = 5 cm
300
AB
 tan 300
BC
B
C
5 3
5
1

BC
3
BC  5 3 cm
By using Pythagoras theorem in Δ ABC , AC2 = AB2 + BC2
 
AC  5  5 3
2
2
2
 25  25.3  25  75  100  10 
2
AC  10 cm
46
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
0
angle bisector
300 30
COB=COA= 300
Length of the chord AB = 2AC=2BC
BC
 Sin300
OB
A
C
0
B
BC
1

6
2
2BC  6
3
6
BC   3
2
Length of the chord AB = 2BC=2(3)=6 cm
47
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
600
3 cm
PQ
 Sin R
PR
1
90
3

 Sin R
6

0
Q
300
R
2
1
 Sin R
2
P  Q  R  1800
P  900  300  1800
Sin300  Sin R
P  1200  1800
R  300
P  1800  1200
PRQ  30
0
P  600
QPR  600
48
Example : 7
If Sin( A  B ) 
1
1
, Cos ( A  B )  , 00  A  B  900
2
2
A > B , Find A and B .
1
Sin( A  B) 
2
Sin( A  B)  Sin 30
A  B  30
0
1
Cos ( A  B ) 
2
Cos( A  B)  Cos 600
A  B  600
A B
  A B
  300  600
0
2 A  90
0
45
90
A
 450

2
A  B  60
0
450  B  600
B  600  450
B  150
49
Exercise 11.2
1. Evaluate the following
(i) sin 450 +cos 450
sin 450 +cos 450 
1
1
11
2




2
2
2
2

2. 2


2
2
cos 450
(ii)
sec 300 +cosec600
1
2
1
1
cos 450
1
3
3
2
2




.

0
0
2
2
22
4
sec 30 +cosec60
2 4
4 2

3
3
3
3
50
Exercise 11.2
Sin300  tan 450  cos ec600
(iii )
cot 450  cos 600  sec 300
1 1
2
 
0
0
0
Sin30  tan 45  cos ec60
2 1
3

0
0
0
1 1
2
cot 45  cos 60  sec 30
 
1 2
3
32 34
3 34
2
3


1
3 34
2 3 3  4
2 3
2
0
2
0
2
0
(iv) 2tan 45 +cos 30 - sin 60 2
 3
2tan 45 +cos 30 - sin 60  2 1   
 2 
3 3
 2 1        2
4 4
2
0
2
0
2
0
2
51
Exercise 11.2
Sec2 600 - tan 2 600
(v )
Sin 2 300 + Cos 2 300
 
 2  3
Sec 2 600 - tan 2 600

2
2
0
2
0
2
Sin 30 + Cos 30
1  3

  
2  2 
2
2
 2  3
43
1
1
1



 1
2
2

1 3
1 3
4
1
 3
1


  

4 4
4
4
2
2


2.Choose the right option and justify your choice.
(
c )
2 tan 2 300
2
2
 
(i )
1  tan 2 450
(a) Sin 600
(b) Cos 600
(c) tan 300
(d) Sin 300
 1   2   2 
2
 
 
 2 1 1
0
2 tan30
3
3
3




0

.


tan
30
2
1  tan 2 450 1  12
11
3 2
3
1
52
Exercise 11.2
2.Choose the right option and justify your choice.
1- tan 2 450
(ii )
1+ tan 2 450
(a) tan
(b) 1
900
d
(
)
(c) Sin 450
(d) 0
1  1
1- tan 45
1 1
0



0
2
2
0
1+ tan 45
11
2
1  1
2
2
0
0
2tan30
(iii )
2
0
1-tan 30
(a) Cos
600

2

2tan300


1-tan 2 300

1 

1 

3
(b) Sin
600
c
)
(c) tan
600
(
 2   2  
    
 3  3 

2
1  1   1  3 1
 

3
 3
3
(d) Sin 300
2 

3. 3
3 2 3
 . 
 3  tan 600
2
3 2
3
3
53
Exercise 11.2
3.Evaluate Sin 600cos 300 + cos 600 Sin 300 . What is the value of
Sin ( 600 + 300 ). What can you conclude ?
Solution : Sin 600cos 300 + Cos 600 Sin 300
1
3
3 1 1
3 1 3 1
4

.
 .   

 1 ………….1
2
2
2 2
4 4
4
4
Sin ( 600 + 300 ) = Sin 900 =1 ………….2
From 1 and 2
Sin ( 600 + 300 ) = Sin 600cos 300 + Cos 600 Sin 300
Sin ( A + B ) = Sin A cos B + Cos A Sin 300
54
Exercise 11.2
4. Is it right to say Cos(600 + 300 ) = cos 600 Cos 300 – Sin 600 Sin 300
Solution : Cos600 cos 300 – Sin 600 Sin 300
1
3
3 1
3
3
 .

. 

 0 ………….1
2 2
2 2
4
4
Cos ( 600 + 300 ) = Cos 900 = 0 ………….2
From 1 and 2
Cos ( 600 + 300 ) = Cos 600cos 300 – Sin 600 Sin 300
Therefore it is right to say
Cos(600 + 300 ) = cos 600 Cos 300 – Sin 600 Sin 300
55
Exercise 11.2
5. In a right angle triangle ΔPQR, right angle is at Q and PQ = 6 cm ,
RPQ = 600 . Determine the lengths of QR and PR .
Solution : Given that right angle at Q in ΔPQR
P
and PQ = 6 cm , RPQ = 600
For angle P in ΔPQR
0
QR
 Tan 600
PQ
QR
 3
6
QR  6 3 cm
QR
 Sin 600
PR
6 cm
60
900
Q
6 3 cm
R
6 3
3

PR
2
6
1

PR
2
PR  12 cm
56
Exercise 11.2
6. In ΔXYZ, right angle is at Y , YZ = x and XY = 2x then determine
YXZ and YZX
(2x)2 = (x)2 + YZ2
Tan 600  Tan X
4x2 = x2 + YZ2
YZ2 = 4x2 – x2
YZ2 = 3x2
YZ 
2


3x
X  60

YXZ  600
For angle Z in ΔXYZ
3x
 Tan X
x
3  Tan X
600
90
Y
0
300
3x
Z
2
YZ  3x
For angle X in ΔXYZ
YZ
 Tan X
XY
0
X
x
Solution : In ΔXYZ , right angle is at Y ,
XZ = 2x , XY = x
By using Pythagoras theorem XZ2 = XY2 + YZ2
XY
 Tan Z
YZ
1
x
 Tan Z
3x
1
 Tan Z
3
Tan 300  Tan Z
Z  300
YZX  300
57
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 (600 + 300 ) = Sin 900 = 1
RHS Sin A + Sin B = Sin
600
+ Sin
300
3 1
3 1
 
=
2 2
2
LHS  RHS
It is not right to say that Sin (A+B) = Sin A + Sin B
58
Think – Discuss
For which value of acute angle Cos  Cos  4 Is true ?
1  Sin 1  Sin
For which value of 00    900, above equation is not defined?
Cos
Cos
Solution :
2Cos  1

4
1  Sin 1  Sin
Cos (1  Sin )  Cos (1  Sin )
1
4
Cos 
(1  Sin )(1  Sin )
2
Cos  Cos .Sin  Cos  Cos .Sin
4
2
2
(1)  (Sin )
2Cos
4
2
cos 
2Cos
 4
2
1  Sin 
1
2
2 Cos
 4
Cos . Cos
Cos  Cos600
  600
For = 600 the given
statement is true .
59
Trigonometric Ratioes of complementary Angles
Two angles are said to be complementary , if their sum is equal to 900
BC
Tan x 
AB
AC
Cosec x 
BC
AC
Sec x 
AB
AB
Cot x 
BC
A  C  900
C  90  A
0
Let
C
x
A
90 0
Adjacent side to x
A  x
Opposite side to x
Adjacent side to 900 - x
BC
Sin x 
AC
AB
Cos x 
AC
B
Opposite side to 900 - x
C  900  x
AB
Sin(90 - x) 
 Cos x
AC
BC
0
Cos (90 - x) 
 Sin x
AC
0
AB
Tan (90 - x) 
 Cot x
BC
0
AC
Cosec (90 - x) 
 Sec x
AB
0
Sec (900 - x) 
AC
 Co sec x
BC
BC
Cot (90 - x) 
 Tan x
AB
0
60
Sin (90  A)  Cos A
0
Cos (90  A)  Sin A
0
Tan (90  A)  Cot A
0
Cot (90  A)  Tan A
0
Sec (90  A)  Co sec A
0
Cosec (90  A)  Sec A
0
61
Example: 8
Sec350
Evaluate
Co sec 550
0
Sec
35
Solution:
Co sec 550
Sec350

Co sec(900  350 )
1
0
Sec 35

0
Sec 35
( Co sec (90  A)  Sec A )
1
62
Example : 9
If Cos 7 A  Sin ( A  60 )
Where 7A is an acute angle, Find the value of A .
0
Solution : Given Cos 7 A  Sin ( A  6 )
Sin (900  7 A)  Sin ( A  60 )
Since 7A is an actute angle, ( 900 – 7A ) and ( A – 60 ) are also actute
(90  7 A)  ( A  6 ) ( Sin A  Sin B  A  B)
0
0
90  7 A  A  6
0
0
90  6  A  7 A
0
0
960  8A
12
96
A
8
A  12
0
0
63
Example : 10
If Sin A = Cos B , then Prove that A + B = 900.
Given that Sin A = Cos B
Solution :
SinA  Sin(900  B)
(
Cos B  Sin(900  B) )
( Sin A  Sin B  A  B)
A  900  B
A+B  900
Example : 11 Express Sin 810 + Tan 810 in terms of trigonometric
ratioes of angle between 00 and 450
Sin 81  Cos(90  81 )  Cos 9
0
0
0
0
Tan 81  Cot (90  81 )  Cot 9
0
0
0
0
Sin 81 +Tan 81  Cos 9  Cot 9
0
0
0
64
0
Example : 11
If A,B and C are interior angles of right angle triangle ABC then Show that
BC
A
Sin
 Cos
2
2
Solution :
Given A,B and C are interior angles of right triangle ABC then
A  B  C  180
90
A  B  C 180

2
2
A BC

 900
2
2
BC
A
0
 90 
2
2
On taking Sin ration on both sides
BC
A
0
Sin
 Sin (90  )
2
2
BC
A
Sin
 Cos
2
2
65
Exercise 11.3 1. Evaluate
tan 360
0
0
0
0
(
iii
)
Co
sec31

Sec
59
(i )
(
ii
)
Cos
12

Sin
78
cot 540
(iv) Sin 150.Sec 750
(v) Tan 260 .Tan 640
tan 360
(i )
cot 540
Solution :
1
0
tan 36
tan 360


1
0
0
0
cot(90  36 )
tan 36
(ii ) Cos120  Sin780  Cos(90  780 )  Sin780
=Sin 780  Sin 780  0
(iii) Co sec31  Sec 59  Co sec31  Sec (90 - 31 )
 Co sec 310  Co sec 310  0
0
0
0
0
0
0
0
0
=
Sin
15
.Sec
(90
-15
)
(iv) Sin 15 .Sec 75
0
0
1
= Sin 15 .Cosec 15 = Sin 15 . Sin 150  1
0
0
0
0
0
0
0
0
(v) Tan 26 .Tan 64 =Tan 26 .Tan (90 - 26 )
1
=Tan 26 .Cot 26 = Tan 26 .
1
0
Tan 26
0
0
0
66
Exercise 11.3
2. Show that
(i) tan 48 tan16 tan 42 tan 74  1
0
0
0
0
(ii)Cos 36 Cos 54  Sin 36 Sin 54  0
0
Solution:
0
0
0
(i ) LHS tan 480 tan160 tan 420 tan 740
 tan 480 tan160 tan(900  480 ) tan(900  160 )
 tan 480 tan160 Cot 480 Cot160
1
1
0
0
 tan 48 tan16
 1 RHS
0
0
tan 48 tan16
LHS = RHS
(ii) LHS Cos 360Cos 540  Sin 360 Sin 540
= Cos 360Cos (900  360 )  Sin 360 Sin(90  360 )
= Cos 360Sin 360  Sin 360Cos 360
0
0
= Cos 36 Sin 36  Cos 36 Sin 36  0 RHS
0
0
LHS = RHS
67
Exercise 11.3 Another Method of Solution
(ii)Cos 360Cos 540  Sin 360 Sin 540  0
(ii) LHS Cos 360Cos 540  Sin 360 Sin 540
 Cos (90  540 )Cos(90  360 )  Sin 360 Sin 540
Solution:
 Sin 54 Sin 36  Sin 36 Sin 54
0
0
0
0
 Sin 54 Sin 36  Sin 54 Sin 36
0
0
0
0
 0 RHS
LHS = RHS
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)
( 900 – 2A) = (A – 180)
tan A  Cot (900  A)
Cot A = Cot B AÆÿ$$¯èþ A=B
900 +180 = A + 2A
3A = 1080
36
108
A
 360
3
68
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
Cot (900  A)  cot B
900  A  B
tan A  cot(900  A)
CotA  CotB  A  B
90  A  B
0
A  B  90
69
Exercise 11.3
5. If A,B and C are interior angles of a triangle ABC, then show that
A B
C
tan
 Cot
2
2
Solution : Given that A,B and C are interior angles of a triangle ABC
A  B  C  180
90
A  B  C 180

2
2
A B C
  900
2
2
A B
C
0
 90 
2
2
By taking tan ratio on both sides
A B
C
0
tan
 tan( 90  )
2
2
A B
C
tan
 cot
2
2
70
Exercise 11.3
6. Express Sin 750 + cos 650 in terms trigonometric
ratioes of angles between 00 and 450
Sin 75  Sin(90  15 )  Cos 15
0
0
0
0
cos 65  cos(90  35 )  Sin 35
0
0
0
0
Sin 75  cos 65  Cos 15  Sin 35
0
0
0
0
71
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
BC 2  AB 2 AC 2

2
AC
AC 2
2
1
2
90 0
B
2
A
BC
AB
AC


2
2
2
AC AC
AC
BC 2 AB 2

1
2
2
AC
AC
2
2
 BC   AB 

 
 1
 AC   AC 
 Sin A
2
  Cos A   1
2
Sin A  Cos A  1
2
2
required trigonometric identity .
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 AB2
BC  AB
AC

2
2
AB
AB
2
2
1
2
2
BC
AC
1 
2
2
AB
AB
2
B
2
BC 2 AB
AC 2


2
2
2
AB
AB
AB
2
90 0
 BC 
 AC 

 1  

 AB 
 AB 
2
A
2
2
 AC 
 BC 




 1
 AB 
 AB 
 Sec A   tan A
2
2
1
Sec A  tan A  1
2
2
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 BC2
BC 2  AB 2
AC 2

2
BC
BC 2
BC 2 AB 2 AC 2


2
2
BC
BC
BC 2
1
BC 2 AB 2 AC 2


2
2
BC
BC
BC 2
2
2
AB
AC
1

2
BC
BC 2
2
 AB 
 AC 
1 
 

BC
BC




2
90 0
B
2
A
2
 AC   AB 

 
 1
 BC   BC 
 Co sec A   Cot A
2
2
1
Co sec A  Cot A  1
2
2
required trigonometric identity
74
Sin 2 A  Cos 2 A  1
Sin A  Cos A  1
Sin A  1  Cos A
Cos A  1  Sin A
2
2
2
2
2
2
Cos A  1  Sin2 A
Sin A  1  Cos 2 A
Sec A  tan A  1
Sec 2 A  1  tan 2 A
Sec 2 A  tan 2 A  1
2
2
tan A  Sec A  1
Sec A  1  tan A
tan A  Sec2 A 1
2
2
2
Co sec A  Cot A  1
2
2
Co sec A  Cot A  1
2
2
Co sec A  1  Cot A
Cot A  Co sec A  1
Co sec A  1  Cot A
Cot A  Co sec2 A 1
2
2
2
2
2
75
15 then finc Cos A
Do this (i ) If Sin A 
17
2
Cos A  1  Sin A
2


 15  
1 

17


2
64
 8 
 

289
17


1
225
289  225

289
289
8

17
5 then find Sec x
(ii ) If tan x 
12
2
25
5


2

1


1



Sec x  1  tan x
 12 
144

144  25
144
2
169
13
 13 

   
 12 
144
12
25
then find cot 
(iii ) If Cosec  
7
2
 25 
2

Cot  Co sec  1
  1 
 7 
 24 
  
 7 
2
625  49
625
576

1

49
49
49
242

72
24

7
76
Try This Evaluate the following and Justify your answer.
0
0
0
0
(
ii
)
Sin
5
Cos
85

Cos
5
Sin
85
Sin 15  Sin 75
(i)
0
0
0
0
2
0
2
0
(
iii
)
Sec
16
Co
sec
74

Cot
74
tan
16
Cos 36  Cos 54
2
0
2
0
Sin 2150  Sin 2 750
(i)
Cos 2 360  Cos 2 540
Sin2150  Sin2 (900  150 )

Cos 2 360  Cos 2 (900  360 )
 Sin(90   )  cos 
Sin2150  Cos 2150

Cos 2 360  Sin2 360
1

1

1
Cos (90   )  Sin  
 Sin2  Cos 2  1

Cos 2  Sin 2  1
(ii) Sin 50Cos 850  Cos 50 Sin 850
= Sin 50Cos (900  50 )  Cos 50 Sin (900  50 )
= Sin 5 Sin 5  Cos 5 Cos 5
0
0
0
0



Cos (90   )  Sin  
Sin(90   )  cos 
Sin 2  Cos 2
= Sin 2 50  Cos 2 50 = 1
(iii) Sec 160Co sec 740  Cot 740tan 160
 1
= Sec 160 Co sec (900  160 )  Cot (90  160 ) tan 160
= Sec 160 Sec 160  tan 160 tan 160
= Sec 16  tan 16 = 1
2
0
2
0

Sec2  tan 2  1
77
Example : 13
Show that Cot  + tan  = Sec  Cosec 
LHS Cot  + tan 
Cot  
Cos 
Sin 
=
+
Sin 
Cos 
Cos 2  Sin2
=
Sin  .Cos 
tan  
Cos 
Sin 
Sin 
Cos 
( Cos2  Sin2  1)
1
1
1
1
1
=
=
.
=
.
Sin  Cos 
Cos  Sin 
Sin  .Cos 
Sec =
1
Cos 
Cosec =
 Sec Co sec RHS LHS  RHS
Example : 14 Show that tan2  + tan4  = Sec4  – sec2 
LHS tan 4   tan 2 
 tan 2  tan 2   tan 2  .1)
 tan 2  (tan 2   1)
 ( Sec 2  1).Sec 2
 ( Sec 2 .Sec 2  Sec2 .1)
 Sec   Sec  RHS
4
2
1
Sin 
tan 2   Sec 2  1
Sec 2  1  tan 2 
LHS  RHS
78
Example : 15
Show that 1  Cos   Co sec  Cot
1  Cos 
1  Cos 
LHS
1  Cos 

1  Cos  (1  Cos  )

1  Cos  (1  Cos  )
(1  Cos  ) 2
12  Cos 2
(1  Cos  ) 2

1  Cos 2
(a  b)(a  b)  a 2  b2
1  Cos 2  Sin 2
(1  Cos  ) 2

Sin 2
 1  Cos  
 

Sin



2
1  Cos 

Sin 
1
Cos 


Sin  Sin 
Co sec  
Cot  
1
Sin 
Cos 
Sin 
 Co sec   Cot  RHS
LHS  RHS
79
Exercise :11.4
1. Evaluate the following
(i ) (1  tan   Sec  )(1  Cot   Cosec  )
(ii) (Sin   Cos  )2  (Sin   Cos  )2
(iii ) (Sec2  1)(Cose2  1)
Solution: (i ) (1  tan   Sec  )(1  Cot   Cosec  )
Sin 
tan  
Cos 
1
Sec  
Cos 
Cos 
Cot  
tan 
1
Cosec  
Sin 
1  Cos 
1   Cos   Sin   1  Sin  Cos   1 
 Sin 
 1 

1





  
Cos

Sin



Cos

Cos

Sin

sin




  Cos   Sin  2  12


Cos  .Sin 

Cos





  Sin2  2Cos  sin   1
Cos  . Sin 
2
1  2Cos  sin   1  1  2Cos  sin   1  2 Cos  sin 

Cos  . Sin 
Cos  . Sin 
Cos  . Sin 
2
80
1. Evaluate the following
Exercise :11.4
(ii) (Sin   Cos  )2  (Sin   Cos  )2
(iii ) (Sec2  1)(Cose2  1)
Solution: (ii) (Sin   Cos
 )2  (Sin   Cos  )2
(a  b)2  a 2  b2  2ab
(a  b)  a  b  2ab
2
2
2
 Sin 2  Cos 2  2Sin Cos  Sin 2  Cos 2  2Sin Cos 
 (Sin   Cos  )  (Sin   Cos  )
2
2
2
2
Sin 2  Cos 2  1
 11  2
(iii ) (Sec2  1)(Cose2  1)
( Sec   1  tan  )
2
2
( Cose2  1  cot 2  )
 tan  .cot 
2
2
1
 tan  .
tan 2 
2
1
81
Exercise : 11.4 2. Show that  Co sec   Cot 
LHS  Co sec   Cot 
Cos  
 1



 Sin  Sin  
1  Cos  


2
2
Sin 2
1  Cos  


Cosec  
1
Sin 
 1  Cos  


 Sin  
2
1  Cos 

1  Cos 
Cos 
Cot  
tan 
1  Cos  


2
Sin



2
Sin 2  1  Cos 2
1  Cos  


2
1  Cos 2

2
2
12  Cos 2
2
a 2  b2  (a  b)(a  b)
1  Cos   1  Cos  
1  Cos   1  Cos  
1  Cos  


1  Cos  
RHS
LHS  RHS
82
Exercise : 11.4
3. show that
LHS
1  SinA
 SecA  TanA
1  SinA
1  SinA
1  SinA
1  SinA (1  SinA)
1  SinA (1  SinA)




(1  SinA)
12  Sin 2 A
(1  SinA) 2
1  Sin 2 A
(1  SinA) 2
Cos 2 A
 1  Sin A 


Cos
A


1  Sin A

Cos A

(a  b)(a  b)  a 2  b2
2
( 1  Sin2 A  Cos 2 A)

1
Sin A

Cos A Cos A
2
(
1
 Sec A)
Cos A
(
Sin A
 Tan A)
Cos A
 Sec A  Tan A RHS
LHS  RHS
83
Exercise : 11.4
1  Tan2 A
2
4. Show that

Tan
A
2
Cot A  1
1  Tan 2 A
LHS
Cot 2 A  1
1  Tan 2 A

1
1
2
Tan A
Cot 2 A 
1  Tan 2 A
1

1  Tan 2 A
Tan 2 A
1  Tan2 A Tan2 A

.
1
1  Tan2 A
 Tan A RHS
2
LHS  RHS
1
Tan 2 A
Another Method
1  Tan 2 A
LHS
Cot 2 A  1
Sin A
Cos A
Cos A
Cot A 
Sin A
Tan A 
Sin 2 A
1
2
Cos
A

Cos 2 A
1
2
Sin A
Cos 2 A  Sin 2 A
2
Cos
A

Cos 2 A  Sin 2 A
Sin 2 A
Cos 2 A  Sin2 A
Sin2 A

.
2
Cos A
Cos 2 A  Sin2 A
Sin 2 A

Cos 2 A
 Tan 2 A RHS
LHS  RHS
Sin A
 Tan A
Cos A
84
Exercise : 11.4
1
5. Show that
 Cos  tan  .Sin
Cos
1
LHS
 Cos  1  Cos 1  Cos 2
2
2
(
1

Cos


Sin
)

Cos
Cos
1
Cos
Sin 2
Sin


.Sin
Cos
Cos
Sin
 tan 
Cos
 tan  .Sin RHS
LHS  RHS
6. Simplify SecA(1  SinA)( SecA  tan A)
SecA(1  SinA)( SecA  tan A)
= (SecA.1  SecA.SinA)( SecA  tan A)
1
= (SecA 
.SinA)( SecA  tan A)
CosA
SinA
= (SecA 
)( SecA  tan A)
CosA
= (SecA  tan A)( SecA  tan A)
= Sec A  tan A
2
2
1
(
(
1
SecA 
)
CosA
SinA
 tan A
CosA
(a  b)(a  b)  a 2  b2
Sec 2 A  tan 2 A  1)
85
Exercise : 11.4
7.Pr ove that (SinA  Co sec A)2  (CosA  SecA) 2  7  tan 2 A  Cot 2 A
LHS (SinA  Co sec A)  (CosA  SecA)
2
(a  b)2  a 2  b2  2ab
2
 (Sin2 A  Co sec2 A  2SinA.Co sec A)  (Cos 2 A  Sec 2 A  2CosA.SecA)
Se c A 
1
Co sec A 
sin A
1
Cos A
1
1
2
2
 ( Sin A  Co sec A  2 SinA .
)  (Cos A  Sec A  2 CosA .
)
SinA
CosA
2
2
 ( Sin2 A  Co sec2 A  2)  (Cos 2 A  Sec 2 A  2)
Sec 2 A  1  tan 2 A
Co sec2 A  1  Cot 2 A
 (Sin2 A  1  Cot 2 A  2)  (Cos 2 A  1  Tan2 A  2)
 Sin A  Cos A  Tan A  Cot A  6
2
2
2
2
( Sin2 A  Cos 2 A  1)
 1  Tan 2 A  Cot 2 A  6
 7  Tan 2 A  Cot 2 A
LHS  RHS
RHS
86
Exercise : 11.4
8. Simplify (1  Cos  )(1  Cos  )(1  Cot  )
2
(1  Cos  )(1  Cos  )(1  Cot 2 )
(a  b)(a  b)  a 2  b2
1  Cot   Co sec 
 (12  Cos 2 )(Co sec2  )
1
2
1
 (1  Cos  )(
)
Co
sec


sin 
Sin 2
1
2
 Sin  .
2
2
2
(
1

Cos


Sin
)
Sin 
2
2
1
9. If Secθ+tanθ = p then what is the value of Sec  tan  ?
Given that Sec  tan   p
a 2  b2  (a  b)(a  b)
Sec   tan   1
( Sec   tan  )( Sec   tan  )  1
2
2
p ( Sec   tan  )  1
1
Sec   tan  
p
87
Exercise : 11.4
10. If Co sec   Cot  k then prove that
k 2 1
Cos  2
k 1
Co sec  Cot  k ............1
2
2
2
2
a

b
 (a  b)(a  b)
Co sec   Cot   1
(Co sec   Cot  )(Co sec   Cot  )  1
(Co sec   Cot  )( k )  1
1 ............ 2
Co sec   Cot  
k
Adding 1 and 2
1
Co sec   Cot 2Co sec   Cot  k 
k
k 1
2Co sec  
k
k 2  1.............. 3
Co sec 
2k
Subtracting 2 from 1
k 2 1
2Cot 
k
2
k 1
Cot 
.............. 4
2k
Dividing 4 by 3
k 2 1
Cot
k
 2
k 2 1
Co sec 
2k
Cos
2
k
Sin   1
1
k 2 1
Sin
Cos
k 2 1
 2
1
k 1
k 2 1
1 Cos  2
 Co sec  Cot    Co sec  Cot   k 
k 1
k
k 2 1
Co sec  Cot  Co sec  Cot 
88
k
Exercise : 11.4
10. If Co sec   Cot  k then prove that
k 2 1
Cos  2
k 1
Another method
Given that Co sec  Cot  k
k 2 1
RHS
k 2 1
2
Co sec   Cot   1

=
2
 Co sec  Cot   1
Cot 
Co sec   Cot   2Co sec  .Cot   1

=
Co sec   Cot   2Co sec .Cot   1
2
2
2
2
( Co sec2   1  Cot 2 )
( Cot 2  Co sec2   1)
1  Cot 2  Cot 2  2Co sec .Cot  1
=
Co sec2   Co sec2   1  2Co sec .Cot  1
2Cot 2  2Co sec  .Cot
=
2Co sec2   2Co sec  .Cot
2 Cot (Cot  Co sec  )
=
2 Co sec  (Cot  Co sec  )
Cot
=
Co sec 
Cos
sin 
1
Co sec  
sin 
Cos
= Sin
1
Sin
Cos
=
1
= Cos LHS
LHS  RHS
89
Optional Exercise
cot   cos  cos ec  1
1. Prove that

cot   cos  cos ec  1
cot   cos 
LHS
cot   cos 
cos 
 cos  .1
= Sin
cos 
 cos  .1
Sin
1
Cos (
 1)
Sin
=
1
Cos (
 1)
Sin
1
1

= Sin 1
1
1

Sin 1
Cot 
Cos
sin 
1  Sin
= Sin
1  Sin
Sin
1  Sin
=
1  Sin
RHS
LHS= RHS
90
Optional Exercise
Sin  cos   1
1
2.Prove that

Sin  cos   1 Sec  tan 
Sin  cos   1
Sec 2  tan 2   1
LHS
Sin  cos   1
( Sec  tan  ).1  ( Sec 2  tan 2  )

Dividing Numeratior and Denominator by Cos
Sin  cos   1
Cos

Sin  cos   1
Cos
Sin  cos 
1


 cos  cos  cos 
Sin  cos 
1


cos  cos  cos 
tan   1  Sec

tan   1  Sec
Sec  tan   1

tan   1  Sec

tan   1  Sec
( Sec  tan  ).1  ( Sec  tan  )( Sec  tan  )
tan   1  Sec
( Sec  tan  )(1  ( Sec  tan  ))

tan   1  Sec
( Sec  tan  ) (1  Sec  tan  )

(1  Sec  tan  )
Sin
cos 
 tan 
1
 Sec
cos 
 Sec  tan 
 Sec  tan  .
Sec  tan 
Sec  tan 
Sec 2  tan 2 

Sec  tan 

1
Sec  tan 
RHS
91
LHS= RHS
Optional Exercise
1
3. Prove that  Co sec A  SinA SecA  CosA  
tan A  CotA
1
LHS  Co sec A  SinA SecA  CosA 
1
CosA.SinA
1
S ec A 
CosA
1
Co sec A 
SinA
 1
 1


 SinA  
 CosA 
 SinA
  CosA

1  Sin2 A  Cos 2 A
1

Sin2 A  Cos 2 A
CosA.SinA
SinA   1
CosA 
 1





SinA
1
CosA
1



 1  Sin 2 A   1  Cos 2 A 



SinA
CosA



1  Sin2 A  Cos 2 A
1  Cos 2 A  Sin 2 A
 Cos 2 A   Sin 2 A 



SinA
CosA



CosA. CosA SinA. SinA

.
SinA
CosA
 CosA.SinA


1
Sin2 A
Cos 2 A

CosA.SinA CosA.SinA
1
SinA. SinA
CosA .CosA

CosA. SinA
CosA .SinA
SinA
 tan A
cos A
1

tan A  cot A
CosA
 CotA
SinA
RHS
LHS= RHS
92
1+SecA
Sin2 A
4. Prove that

SecA
1  CosA
1+SecA
LHS
1+CosA 1  CosA
SecA
Optional Exercise

SecA 
1
cos A
1
1
+
 1 CosA
1
CosA
CosA+1
 CosA
1
CosA
CosA+1

1
1+CosA

1
1
.
1  CosA
a 2  b2  (a  b)(a  b)
12  Cos 2 A

1  CosA
1  Cos 2 A

1  CosA
Sin 2 A

1  CosA
RHS
LHS= RHS
93
Optional Exercise  1  tan 2 A  1  tan A 2


2
5. Show that 


tan
A
 

2
2
 1  Cot A   1  CotA 
2
1

tan
A


 1  tan A 
LHS
LHS1 
2


2
1

CotA


 1  Cot A 
1  tan 2 A  Sec 2 A
1  Cot 2 A  Cosec 2 A
Sec 2 A

Cosec 2 A
1
SecA 
cos A
SinA
tan A 
cos A
CotA 
CosA
SinA
2
Co sec A 
1
SinA
1
2
Cos
A

1
Sin 2 A
1
Sin 2 A

.
2
Cos A
1
Sin 2 A
SinA

 tan A
2
Cos A
cos A
 tan 2 A RHS
 SinA   CosA  SinA 2
 1  cos A  

cos
A




SinA

CosA
CosA

 1
 
SinA

 SinA  
 SinA  CosA 


cos
A

SinA  CosA 


SinA


2
 1 
2
1
SinA

 cos A   
.




cos A 1 
1



 SinA 
2
2
 SinA   tan A 2


 
 cos A 
 tan 2 A RHS
LHS1 =LHS2 = RHS
94
Optional Exercise
 SecA  1   1  CosA 
6. Prove that 


 SecA  1   1  CosA 
 SecA  1 
LHS 

SecA

1


1
SecA 
cos A
 1

 CosA  1 
= 

1

1 
 CosA

 1  CosA
 CosA
= 
1  CosA

 CosA
 1  CosA 
=

1

CosA







LHS= RHS
RHS
95
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