Trigonometric ratios and identities 1

ppt for class XI

1. Mathematics

2. Trigonometric ratios and Identities
Session 1

3. Topics
Transformation of Angles
Compound Angles
Definition and Domain and Range
of Trigonometric Function
Measurement of Angles

4. Measurement of Angles
O A
B
OA InitialRay-
uuur
OB Terminal Ray-
uuur
Angle is considered as the figure obtained by
rotating initial ray about its end point.
J001

5. Measure and Sign of an Angle
Measure of an Angle :-
Amount of rotation from initial side
to terminal side.
Sign of an Angle :-
O A
B
q
Rotation anticlockwise –
Angle positive
B’
-q
Rotation clockwise –
Angle negative
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6. Right Angle
O
Y
X
Revolving ray describes one – quarter of a circle then
we say that measure of angle is right angle
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Angle < Right angle Þ Acute Angle
Angle > Right angle Þ Obtuse Angle

7. Quadrants
O
Y
Y’
X’ X
II Quadrant
( , )- +
I Quadrant
( , )+ +
IV Quadrant
( , )+ -
III Quadrant
( , )- -
X’OX – x - axis
Y’OY – y - axis
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8. System of Measurement of Angle
Measurement of Angle
Sexagesimal System
or
British System
Centesimal System
or
French System
Circular System
or
Radian Measure
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9. System of Measurement of Angles
Sexagesimal System (British System)
1 right angle = 90 degrees (=90
o
)
1 degree = 60 minutes (=60’)
1 minute = 60 seconds (=60”)
Centesimal System (French System)
1 right angle = 100 grades (=100
g
)
1 grade = 100 minutes (=100’)
1 minute = 100 Seconds (=100”)
J001
Is 1 minute of
sexagesimal
1 minute of
centesimal ?
=
NO

10. System of Measurement of Angle
Circular System
J001
O
r
r
r
A
B
1
c
If OA = OB = arc AB
c
Then AOB 1radian( 1 )Ð = =

11. System of Measurement of Angle
Circular System
O A
C
B
1
c
AOC arcAC
AOB arcACB
Ð
=
Ð
Q
1radian r
2rightangles r
Þ =
p
2rightangles radianÞ =p
J001
180 radianÞ = p
o

12. Relation Between Degree Grade
And Radian Measure of An Angle
0 g
D G 2C
90 100
= =
p
OR
0 g
D G C
180 200
= =
p
J002

13. Illustrative Problem
Find the grade and radian measures
of the angle 5
o
37’30”
Solution
o' o
30 30 1
30"
60 60 60 120
æ öæ ö æ ö
= = =
ç ¸ ç ¸ç ¸
´è ø è øè ø
o
37
and37'
60
æ ö
=
ç ¸
è ø
o o
o 37 1 45
5 37'30" 5
60 120 8
æ ö æ ö
\ = + + =
ç ¸ ç ¸
è ø è ø
J002
We know that
D G 2C
90 100
= =
p

14. Illustrative Problem
Find the grade and radian measures of the
angle 5
o
37’30”
g
10
G D
9
æ ö
Þ = ´
ç ¸
è ø
( )
g g
g10 45 225
12.5 Ans
9 8 18
æ ö æ ö
= ´ = =
ç ¸ ç ¸
è ø è ø
c
and R D
180
pæ ö
= ´
ç ¸
è ø
c
45
radian Ans
180 8 32
p pæ ö
= ´ =
ç ¸
è ø
Solution
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15. Relation Between Angle Subtended

by an Arc At The Center of Circle
O A
C
1
c
q
B
Arc AC = r and Arc ACB = 
AOC arcAC
AOB arcACB
Ð
=
Ð
Q
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1radian r
Þ =
q l
r
Þ q=
l

16. Illustrative Problem
A horse is tied to a post by a rope. If
the horse moves along a circular path
always keeping the rope tight and
describes 88 meters when it has traced
out 72
o
at the center. Find the length of
rope. [ Take p = 22/7 approx.].
Solution
P A
B
72
o
Arc AB = 88 m and AP = ?
c
o 2
72 72 rad
180 5
p pæ ö
q= = ´ =
ç ¸
è ø
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arcAB
r AP
q= =
l
Q
2 88 22
AP 70m [ approx.]
5 AP 7
p
= Þ = p=

17. Definition of Trigonometric Ratios
J003
2 2
r x y= +
x
O
Y
X
P (x,y)
M
q
y
r
y
sin
r
x
cos
r
y
tan
x
q=
q =
q=
x
cot
y
r
sec
x
r
cosec
y
q=
q =
q=

18. Some Basic Identities
sin cosec 1 ; n ,n I· q´ q= q ¹ p Î
2 2
sin cos 1· q+ q =
2 2
sec tan 1· q- q =
2 2
cosec cot 1· q- q =
( )
sin
tan ; 2n 1 ,n I
cos 2
q p
· q = q ¹ + Î
q
( )tan cot 1 ; 2n 1 ; n ,n I
2
p
· q´ q= q ¹ + q ¹ p Î
cos
cot ; n ,n I
sin
q
· q = q ¹ p Î
q
( )cos sec 1 ; 2n 1 ,n I
2
p
· q´ q= q ¹ + Î

19. Illustrative Problem
Solution
3
sec tan .cosec= q+ q q
3cosec
sec 1 tan
sec
qæ ö
= q + q
ç ¸
qè ø
( )
3
sec 1 tan cot= q + q q ( )
2
sec 1 tan= q + q
( )
2 2
1 tan 1 tan= + q + q
( )
3
2
2
1 tan= + q
( )
3
2
2
2 e Proved= -
J003
2 2
Iftan 1 e ,provethatq = -
( )
3
3 2
2
sec tan .cosec 2 eq+ q q= -
0
2
pæ ö
< q <
ç ¸
è ø

20. Signs of Trigonometric Function In
All Quadrants
In First Quadrant
x
O
Y
X
P (x,y)
M
q
y
r
Here x >0, y>0,
2 2
r x y= + >0
y
sin 0
r
q = >
x
cos 0
r
q = >
y
tan 0
x
q = >
x
cot 0
y
q = >
r
sec 0
x
q = >
r
cosec 0
y
q = >
J004

21. Signs of Trigonometric Function In
All Quadrants
In Second Quadrant
Here x <0, y>0,
2 2
r x y= + >0
y
sin 0
r
q = >
r
cosec 0
y
q = >
q
XX’
Y
Y’
P (x,y)
x
y
r
x
cos 0
r
q = <
y
tan 0
x
q = <
x
cot 0
y
q = <
r
sec 0
x
q = <
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22. Signs of Trigonometric Function In
All Quadrants
In Third Quadrant
Here x <0, y<0,
2 2
r x y= + >0
r
cosec 0
y
q = >
r
sec 0
x
q = <
q
X’ X
P (x,y)
O
Y’
Y
M
y
sin 0
r
q = <
x
cos 0
r
q = <
y
tan 0
x
q = >
x
cot 0
y
q = >
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23. Signs of Trigonometric Function In
All Quadrants
In Fourth Quadrant
Here x >0, y<0,
2 2
r x y= + >0
y
sin 0
r
q = <
q
XO
P (x,y)
Y’
M
x
cos 0
r
q = >
y
tan 0
x
q = <
x
cot 0
y
q = <
r
sec 0
x
q = >
r
cosec 0
y
q = <
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24. Signs of Trigonometric Function In
All Quadrants
I Quadrant
All Positive
II Quadrant
sin & cosec
are Positive
III Quadrant
tan & cot are
Positive
IV Quadrant
cos & sec are
Positive
X
Y’
X’
Y
O
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ASTC :- All Sin Tan Cos

25. Illustrative Problem
q lies in secondIf cot q =
12
,
5
-
quadrant, find the values of
other five trigonometric function
Solution
12 5
cot tan
5 12
q = - Þ q = -Q
2 2 2 169
sec 1 tan sec
144
q = + q Þ q =Q
( )
13 13
sec sec liesinsecondquadrant
12 12
Þ q = ± Þ q = - q
12
Whichgivescos
13
q = -
13
cosec
5
\ q =
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Method : 1
5 12 5
Thensin tan cos
12 13 13
q = q´ q = - ´- =

26. Illustrative Problem
q lies in secondIf cot q =
12
,
5
-
quadrant, find the values of other five
trigonometric function
Solution
J004
Method : 2
Y
q
XX’
Y’
P (-12,5)
-12
5
r
Here x = -12, y = 5 and r = 13
y 5
sin
r 13
q = =
x 12
cos
r 13
-
q = =
y 5
tan
x 12
q = =
-
r 13
sec
x 12
q = =
-
r 13
cosec
y 5
q = =

27. Functions Domain Range
sinq R [-1,1]
cosq R
[-1,1]
secq
R : (2n 1)
2
pì ü
- q q = +í ý
î þ
R-(-1,1)
cosecq { }R : n- q q = p R-(-1,1)
tanq R : (2n 1)
2
pì ü
- q q = +í ý
î þ
R
cotq
{ }R : n- q q = p R
Domain and Range of Trigonometric
Function
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28. Illustrative problem
Prove that
2
2 (x y)
sin
4xy
+
q =
is possible for real values of x and
y only when x=y
Solution
( )
2
2(x y)
1 x y 4xy
4xy
+
Þ £ Þ + £
2
sin 1q £Q
( ) ( )
2 2
x y 4xy 0 x y 0Þ + - £ Þ - £
But for real values of x and y
( )
2
x y- is not less than zero
( )
2
x y 0 x y Proved\ - = Þ =
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29. Trigonometric Function For Allied
Angles
Trig. ratio-q 90
o
-q90
o
+q180
o
-q180
o
+q360
o
-q360
o
+q
cosq cosqsinq- sinq- cosq - cosqcosq cosq
tanq - tanqcotq- cotq-tanq tanq- tanqtanq
sinq - sinqcosqcosq sinq - sinq- sinqsinq
If angle is multiple of 90
0
then
sin Û cos;tan Û cot; sec Û cosec
If angle is multiple of 180
0
then
sin Û sin;cos Û cos; tan Û tan etc.

30. Trigonometric Function For Allied
Angles
Trig. ratio-q 90
o
-q90
o
+q180
o
-q180
o
+q360
o
-q360
o
+q
secq secqcosecq- cosecq- secq - secqsecq secq
cosecq- cosecqsecq secqcosecq-cosecq- cosecqcosecq
cotq - cotqtanq-tanq -cotq cotq- cotqcotq

31. Periodicity of Trigonometric
Function
Periodicity : After certain value of
x the functional values repeats
itself
Period of basic trigonometric functions
sin (360
o
+q) = sinq Þ period of sinq is 360
o
or 2p
cos (360
o
+q) = cosq Þ period of cosq is 360
o
or 2p
tan (180
o
+q) = tanq Þ period of tanq is 180
o
or p
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If f(x+T) = f(x) " x,then T is called
period of f(x) if T is the smallest
possible positive number

32. Trigonometric Ratio of Compound
Angle
Angles of the form of A+B, A-B,
A+B+C, A-B+C etc. are called
compound angles
(I) The Addition Formula
· sin (A+B) = sinAcosB + cosAsinB
· cos (A+B) = cosAcosB - sinAsinB
( )
tanA tanB
tan A B
1 tanAtanB
+
· + =
-
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( )
sin(A B)
Proof: tan A B
cos(A B)
+
- + =
+
Q
sinAcosB cosAsinB
cosAcosB sinAsinB
+
=
-

33. Trigonometric Ratio of Compound
Angle
J006
sinAcosB cosAsinB
cosAcosB sinAsinB
+
=
-
r r
Dividing N and D bycosAcosB
We get
tanA tanB
1 tanAtanB
+
=
-
Proved
( )
cotBcotA 1
cot A B
cotB cotA
-
· + =
+

34. Illustrative problem
Find the value of
(i) sin 75
o
(ii) tan 105
o
Solution
(i) Sin 75
o
= sin (45
o
+ 30
o
)
= sin 45
o
cos 30
o
+ cos 45
o
sin 30
o
1 3 1 1 3 1
2 22 2 2 2
+
= ´ + ´ =
( )(ii) Ans: 2 3- +

35. Trigonometric Ratio of
Compound Angle
(I) The Difference Formula
· sin (A - B) = sinAcosB - cosAsinB
· cos (A - B) = cosAcosB + sinAsinB
( )
tanA tanB
tan A B
1 tanAtanB
-
· - =
+
Note :- by replacing B to -B in addition
formula we get difference formula
( )
cotBcotA 1
cot A B
cotA cotB
+
· - = -
-

36. Illustrative problem
If tan (q+a) = a and tan (q - a) = b
Prove that
a b
tan2
1 ab
-
a =
-
Solution
( )( ){ }tan2 tana = q + a - q- a
( ) ( )
( ) ( )
tan tan
1 tan tan
q+ a - q -a
=
+ q + a q- a
a b
1 ab
-
=
+

37. Some Important Deductions
· sin (A+B) sin (A-B) = sin
2
A - sin
2
B = cos
2
B - cos
2
A
· cos (A+B) cos (A-B) = cos
2
A - sin
2
B = cos
2
B - sin
2
A
( )
tanA tanB tanC tanAtanBtanC
tan A B C
1 tanAtanB tanBtanC tanCtanA
+ + -
· + + =
- - -

38. To Express acosq + bsinq in the
form kcosf or lsiny
acosq +bsinq
2 2
2 2 2 2
a b
a b cos sin
a b a b
æ ö
= + q + qç ¸
+ +è ø
2 2 2 2
a b
Letcos ,thensin
a b a b
a = a =
+ +
( )
2 2
acos bsin a b cos cos sin sin\ q + q = + q a + q a
( )
2 2
a b cos= + q -a
Similarly we get acosq + bsinq = lsiny
2 2
kcos ,where k a b ,= f = + f = q- a

39. Illustrative problem
7cosq +24sinq
Find the maximum and minimum
values of 7cosq + 24sinq
Solution
2 2
2 2 2 2
7 24
7 24 cos sin
7 24 7 24
æ ö
= + q+ qç ¸
+ +è ø
7 24
25 cos sin
25 25
æ ö
= q + q
ç ¸
è ø
7 24
Letcos sin
25 25
a = Þ a =
( )7cos 24sin 25 cos cos sin sin\ q+ q = q a + q a

40. Illustrative problem
Find the maximum and minimum value of
7cosq + 24sinq
Solution
( )25cos 25cos where= q- a = f f = q- a
1 cos 1- £ f £Q
25 25cos 25Þ - £ f £
\ Max. value =25, Min. value = -25 Ans.

41. Transformation Formulae
· Transformation of product into sum
and difference
· 2 sinAcosB = sin(A+B) + sin(A - B)
· 2 cosAsinB = sin(A+B) - sin(A - B)
· 2 cosAcosB = cos(A+B) + cos(A - B)
Proof :- R.H.S = cos(A+B) + cos(A - B)
= cosAcosB - sinAsinB+cosAcosB+sinAsinB
= 2cosAcosB =L.H.S
· 2 sinAsinB = cos(A - B) - cos(A+B) [Note]

42. Transformation Formulae
· Transformation of sums or difference
into products
C D C D
sinC sinD 2sin cos
2 2
+ -
· + =
C D C D
sinC sinD 2cos sin
2 2
+ -
· - =
C D C D
cosC cosD 2cos cos
2 2
+ -
· + =
C D C D
cosC cosD 2sin sin
2 2
+ -
· - = -
C D D C
cosC cosD 2sin sin
2 2
+ -
· - =
or Note
By putting A+B = C and A-B = D in
the previous formula we get this result

43. Illustrative problem
Prove that
cos8x cos6x cos4x
cot6x
sin8x sin6x sin4x
+ +
=
+ +
Solution
(cos8x cos4x) cos6x
L.H.S
(sin8x sin4x) sin6x
+ +
=
+ +
8x 4x 8x 4x
2cos cos cos6x
2 2
8x 4x 8x 4x
2sin cos sin6x
2 2
+ -
+
=
+ -
+
2cos6xcos2x cos6x
2sin6xsin2x sin6x
+
=
+
2cos6x(cos2x 1)
2sin6x(cos2x 1)
+
=
+
cot6x= Proved

44. Class Exercise - 1
If the angular diameter of the moon
be 30´, how far from the eye can a
coin of diameter 2.2 cm be kept to
hide the moon? (Take p =
approximately)
22
7
A
B
E ( E y e )
r
M o o n

45. Class Exercise - 1
If the angular diameter of the moon be 30´,
how far from the eye can a coin of diameter
2.2 cm be kept to hide the moon? (Take p =
approximately)
22
7
A
B
E ( E y e )
r
M o o n
Solution :-
Let the coin is kept at a distance r
from the eye to hide the moon
completely. Let AB = Diameter of
the coin. Then arc AB = Diameter
AB = 2.2 cm
c c
30 1
30´
60 2 180 360
p pæ ö æ ö æ ö
q = = = ´ =
ç ÷ ç ÷ ç ÷
è ø è ø è ø
o
arc 2.2
radius 360 r
p
\q = Þ =
360 2.2 360 2.2 7
r 252 cm
22
´ ´ ´
= = =
p

46. Class Exercise - 2
Solution :-
Prove that tan3A tan2A tanA =
tan3A – tan2A – tanA.
We have 3A = 2A + A
Þ tan3A = tan(2A + A)
Þ tan3A =
tan2A tanA
1–tan2AtanA
+
Þ tan3A – tan3A tan2A tanA = tan2A + tanA
Þ tan3A – tan2A – tanA = tan3A tan2A tanA (Proved)

47. Class Exercise - 3
If sina = sinb and cosa = cosb, then
–
sin 0
2
a b
=(c)
–
cos 0
2
a b
=(d)
sin 0
2
a+b
=(a) cos 0
2
a+b
=(b)
Solution :-
a= bQsin sin a = bcos cosand
Þ a b= a b=sin – sin 0 and cos –cos 0
– –
2sin cos 0 and –2sin sin 0
2 2 2 2
a b a+b a b a+b
Þ = =
–
sin 0
2
a b
Þ =

48. Class Exercise - 4
Prove that
° ° ° °=
3
sin20 sin40 sin60 sin80
16
LHS = sin20° sin40° sin60° sin80°
Solution:-
1
sin60 [2sin20 sin40 ] sin80
2
= °´ ° ° ´ °
3 1
[cos(40 –20 )–cos(40 20 )] sin80
2 2
= ´ ° ° ° + ° ´ °
3
[cos20 –cos60 ]sin80
4
= ° ° °
3 1
sin80 cos20 – sin80
4 2
é ù
= ° ° ´ °
ê ú
ë û
[ ]
3
2sin80 cos20 –sin80
8
= ° ° °

49. Class Exercise - 4
Prove that
° ° ° °=
3
sin20 sin40 sin60 sin80
16
Solution:-
[ ]
3
sin(80 20 ) sin(80 –20 )–sin80
8
= ° + ° + ° ° °
[ ]
3
sin100 sin60 –sin80
8
= °+ ° °
[ ]
3
sin(180 –80 ) sin60 –sin80
8
= ° ° + ° °
3 3
sin80 –sin80
8 2
é ù
= °+ °ê ú
ê úë û
3
16
=
Proved.

50. Class Exercise - 5
Prove that
n n
n
cosA cosB sinA sinB
sinA –sinB cosA –cosB
A B
2cot , if n is even
2
0, if n is odd
+ +æ ö æ ö
+
ç ÷ ç ÷
è ø è ø
ì +æ ö
ï ç ÷
=í è ø
ï
î
Solution :-
n n
cosA cosB sinA sinB
LHS
sinA–sinB cosA–cosB
+ +æ ö æ ö
= +
ç ÷ ç ÷
è ø è ø
n n
A B A–B A B A–B
2cos cos 2sin cos
2 2 2 2
A B A–B A B A –B
2cos sin –2sin sin
2 2 2 2
+ +æ ö æ ö
ç ÷ ç ÷
= +ç ÷ ç ÷
+ +
ç ÷ ç ÷
ç ÷ ç ÷
è ø è ø
n n
A–B A–B
cot –cot
2 2
é ù é ùæ ö æ ö
= +
ê ú ê úç ÷ ç ÷
è ø è øë û ë û

51. Class Exercise - 5
Solution :-
n n n
A–B A–B
cot (–1) cot
2 2
æ ö æ ö
= +
ç ÷ ç ÷
è ø è ø
n
n n
A–B
2cot , if n is evenA–B
cot 1 (–1) 2
2
0, if n is odd
ì
ïæ ö
é ù= + = íç ÷ ë û
è ø ï
î
Prove that
n n
n
cosA cosB sinA sinB
sinA –sinB cosA –cosB
A B
2cot , if n is even
2
0, if n is odd
+ +æ ö æ ö
+
ç ÷ ç ÷
è ø è ø
ì +æ ö
ï ç ÷
=í è ø
ï
î

52. Class Exercise - 6
The maximum value of 3 cosx + 4
sinx + 5 is
(d) None of these
(a) 5
(b) 9
(c) 7
2 2
3cosx 4sinx 3 4+ = +Q
Solution :-
2 2 2 2
3 4
cosx sinx
3 4 3 4
æ ö
+ç ÷
ç ÷
+ +è ø
3 4
5 cosx sinx
5 5
æ ö
= +
ç ÷
è ø
[ ]5 cosxcos sinxsin= a + a
3 4
Let cos sin
5 5
é ù
a = Þ a =
ê ú
ë û

53. Class Exercise - 6
The maximum value of 3 cosx + 4 sinx + 5
is
Solution :-
3 4
Let cos sin
5 5
é ù
a = Þ a =
ê ú
ë û
5cos(x– )= a
–1 cos(x– ) 1£ a £Q
–5 5cos(x– ) 5Þ £ a £
–5 5 5cos(x– ) 5 10Þ + £ a + £
0 3cosx 4sinx 5 10Þ £ + + £
\ Maximum value of the given expression = 10.

54. Class Exercise - 7
If a and b are the solutions of a
cosq + b sinq = c, then show that
2 2
2 2
a –b
cos( )
a b
a+b=
+
Solution :-
We have … (i)acos bsin cq+ q =
acos c–bsinÞ q = q
2 2 2
a cos (c–bsin )Þ q = q
( )
2 2 2 2 2
a 1–sin c –2bcsin b sinÞ q = q + q
( )
2 2 2 2 2
a b sin –2bcsin (c –a ) 0Þ + q q + =
\abare roots of equatoin (i),

55. Class Exercise - 7
If a and b are the solutions of acosq + bsinq
= c, then show that a+b =
+
2 2
2 2
a –b
cos( )
a b
Solution :-
2 2
2 2
c –a
sin sin
a b
a b =
+
Hence
Again from (i),bsin c–acosq = q
2 2 2
b sin (c–acos )Þ q = q
2 2 2 2 2
b (1–cos ) c a cos –2cacosÞ q = + q q
2 2 2 2 2
(a b )cos –2accos c –b 0Þ + q q + =
\sina and sinb are roots of equ. (ii).

56. Class Exercise - 7
If a and b are the solutions of acosq + bsinq
= c, then show that a+b =
+
2 2
2 2
a –b
cos( )
a b
Solution :-(iv)
\a and b be the roots of equation (i),
\cosa and cosb are the roots of equation (iv).
2 2
2 2
c –b
cos cos
a b
a b =
+
2 2 2 2 2 2
2 2 2 2 2 2
c –b c –a a –b
–
a b a b a b
= =
+ + +
cos( ) cos cos –sin sina + b = a b a bNow

57. Class Exercise - 8
If a seca – c tana = d and b seca
+ d tana = c, then
(a) a
2
+ b
2
= c
2
+ d
2
+ cd
(c) a
2
+ b
2
= c
2
+ d
2
(d) ab = cd
+ = +
2 2
2 2
1 1
a b
c d
(b)

58. Class Exercise - 8
If a seca – c tana = d and b seca
+ d tana = c, then
asec –ctan da a =Q
a csin
– d
cos cos
a
Þ =
a a
Solution :-
bsec dtan ca+ a=QAgain
b dsin
c
cos cos
a
Þ + =
a a
Þ = a ab ccos –dsin ….. ii
a csin dcosÞ = a + a ….. (I)
Squaring and adding
(i) and (ii), we get
2 2 2 2 2 2 2 2
a b c (sin cos ) d (cos sin )+ = a + a + a + a
2cd sin cos 2cd sin cos+ a a - a a
2 2
c d= +

59. Class Exercise -9
2 2 A A
sin –sin –
8 2 8 2
p pæ ö æ ö
+
ç ÷ ç ÷
è ø è ø
The value of
(a) 2 sinA
(c) 2 cosA
1
sinA
2
(b)
1
cosA
2
(d)

60. Class Exercise -9
2 2A A
sin –sin –
8 2 8 2
p pæ ö æ ö
+
ç ÷ ç ÷
è ø è ø
Q
Solution :-
A A A A
sin – sin – –
8 2 8 2 8 2 8 2
ì ü ì üp p p pæ ö æ ö æ ö æ ö
= + + +í ý í ýç ÷ ç ÷ ç ÷ ç ÷
è ø è ø è ø è øî þ î þ
1
sin sinA sinA
4 2
p
= × =
2 2
sin A–sin B sin(A B)sin(A–B)
é ù
= +
ê úë û
Q
2 2 A A
sin –sin –
8 2 8 2
p pæ ö æ ö
+
ç ÷ ç ÷
è ø è ø
The value of

61. Class Exercise -10
If , ,
a and b lie between0 and , then value
a of tan2a is
4
cos( )
5
a+b =
5
sin( – )
13
a b =
4
p
(a) 1
(c) 0
(b)
56
33
(d)
33
56
Solution :- \a and bbetween 0 and ,
p
4
– – and 0
4 4 2
p p p
\ < a b < < a + b <
Consequently, cos(a - b) and sin(a + b) are positive.
2
sin( ) 1–cos ( )a +b = a +b

62. Class Exercise -10
Solution :-
16 3
1–
25 5
= =
2 25 12
cos( – ) 1–sin ( – ) 1–
169 13
a b = a b = =
3 5
tan( ) , tan( – )
4 12
\ a +b = a b =
tan2 tan[( ) ( – )]a = a +b + a b
3 5
tan( ) tan( – ) 56
4 12
3 51–tan( )tan( – ) 33
1–
4 12
+
a + b + a b
= = =
a + b a b
´
If , ,

a and b lie between0 and , then value
a of tan2a is
4
cos( )
5
a+b =
5
sin( – )
13
a b =
4
p