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 Ter minal 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
θ
Rotation anticlockwise –
Angle positive
B’
− θ
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 ⇒ Acute Angle
Angle > Right angle ⇒ 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 (=90o
)
1 degree = 60 minutes (=60’)
1 minute = 60 seconds (=60”)
Centesimal System (French System)
1 right angle = 100 grades (=100g
)
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
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O
r
r
r
A
B
1c
If OA = OB = arc AB
c
Then AOB 1radian( 1 )∠ = =

11. System of Measurement of Angle
Circular System
O A
C
B
1c
AOC arc AC
AOB arc ACB
∠
=
∠
Q
1radian r
2right angles r
⇒ =
π
2right angles radian⇒ = π
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180 radian⇒ = πo

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

13. Illustrative Problem
Find the grade and radian measures
of the angle 5o
37’30”
Solution
o' o
30 30 1
30"
60 60 60 120
    
= = = ÷  ÷ ÷×    
o
37
and37'
60
 
=  ÷
 
o o
o 37 1 45
5 37'30" 5
60 120 8
   
∴ = + + = ÷  ÷
   
J002
We know that
D G 2C
90 100
= =
π

14. Illustrative Problem
Find the grade and radian measures of the
angle 5o
37’30”
g
10
G D
9
 
⇒ = × ÷
 
( )
g g
g10 45 225
12.5 Ans
9 8 18
   
= × = = ÷  ÷
   
c
and R D
180
π 
= × ÷
 
c
45
radian Ans
180 8 32
π π 
= × = ÷
 
Solution
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15. Relation Between Angle Subtended
by an Arc At The Center of Circle
O A
C
1c
θ
B
Arc AC = r and Arc ACB = 
AOC arc AC
AOB arc ACB
∠
=
∠
Q
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1radian r
⇒ =
θ l
r
⇒ θ =
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 72o
at the center. Find the length of
rope. [ Take π = 22/7 approx.].
Solution
P A
B
72o
Arc AB = 88 m and AP = ?
c
o 2
72 72 rad
180 5
π π 
θ = = × = ÷
 
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arc AB
r AP
θ = =
l
Q
2 88 22
AP 70m [ approx.]
5 AP 7
π
= ⇒ = π =

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

18. Some Basic Identities
sin cosec 1 ; n ,n I• θ × θ = θ ≠ π ∈
2 2
sin cos 1• θ + θ =
2 2
sec tan 1• θ − θ =
2 2
cosec cot 1• θ − θ =
( )
sin
tan ; 2n 1 ,n I
cos 2
θ π
• θ = θ ≠ + ∈
θ
( )tan cot 1 ; 2n 1 ; n ,n I
2
π
• θ × θ = θ ≠ + θ ≠ π ∈
cos
cot ; n ,n I
sin
θ
• θ = θ ≠ π ∈
θ
( )cos sec 1 ; 2n 1 ,n I
2
π
• θ × θ = θ ≠ + ∈

19. Illustrative Problem
Solution
3
sec tan .cosec= θ + θ θ
3 cosec
sec 1 tan
sec
θ 
= θ + θ ÷θ 
( )3
sec 1 tan cot= θ + θ θ ( )2
sec 1 tan= θ + θ
( )2 2
1 tan 1 tan= + θ + θ ( )
3
2 21 tan= + θ
( )
3
2 22 e Proved= −
J0032 2
If tan 1 e ,prove thatθ = −
( )
3
3 2 2sec tan .cosec 2 eθ + θ θ = −
0
2
π 
< θ < ÷
 

20. Signs of Trigonometric Function In
All Quadrants
In First Quadrant
xO
Y
X
P (x,y)
M
θ
y
r
Here x >0, y>0, 2 2
r x y= + >0
y
sin 0
r
θ = >
x
cos 0
r
θ = >
y
tan 0
x
θ = >
x
cot 0
y
θ = >
r
sec 0
x
θ = >
r
cosec 0
y
θ = >
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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
θ = >
r
cosec 0
y
θ = >
θ
XX’
Y
Y’
P (x,y)
x
y
r
x
cos 0
r
θ = <
y
tan 0
x
θ = <
x
cot 0
y
θ = <
r
sec 0
x
θ = <
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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
θ = >
r
sec 0
x
θ = <
θ
X’ X
P (x,y)
O
Y’
Y
M
y
sin 0
r
θ = <
x
cos 0
r
θ = <
y
tan 0
x
θ = >
x
cot 0
y
θ = >
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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
θ = <
θ
XO
P (x,y)
Y’
M
x
cos 0
r
θ = >
y
tan 0
x
θ = <
x
cot 0
y
θ = <
r
sec 0
x
θ = >
r
cosec 0
y
θ = <
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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
θ lies in secondIf cot θ =
12
,
5
−
quadrant, find the values of
other five trigonometric function
Solution
12 5
cot tan
5 12
θ = − ⇒ θ = −Q
2 2 2 169
sec 1 tan sec
144
θ = + θ ⇒ θ =Q
( )
13 13
sec sec liesinsec ondquadrant
12 12
⇒ θ = ± ⇒ θ = − θ
12
Whichgivescos
13
θ = −
13
cosec
5
∴ θ =
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Method : 1
5 12 5
Thensin tan cos
12 13 13
θ = θ × θ = − × − =

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

27. Functions Domain Range
sinθ R [-1,1]
cosθ R [-1,1]
secθ R : (2n 1)
2
π 
− θ θ = + 
 
R-(-1,1)
cosecθ { }R : n− θ θ = π R-(-1,1)
tanθ R : (2n 1)
2
π 
− θ θ = + 
 
R
cotθ
{ }R : n− θ θ = π R
Domain and Range of Trigonometric
Function
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28. Illustrative problem
Prove that
2
2 (x y)
sin
4xy
+
θ =
is possible for real values of x and
y only when x=y
Solution
( )
2
2(x y)
1 x y 4xy
4xy
+
⇒ ≤ ⇒ + ≤
2
sin 1θ ≤Q
( ) ( )2 2
x y 4xy 0 x y 0⇒ + − ≤ ⇒ − ≤
But for real values of x and y
( )2
x y− is not less than zero
( )2
x y 0 x y Pr oved∴ − = ⇒ =
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29. Trigonometric Function For Allied
Angles
Trig. ratio -θ 90o
-θ 90o
+θ 180o
-θ 180o
+θ 360o
-θ 360o
+θ
cosθ cosθ sinθ - sinθ - cosθ - cosθ cosθ cosθ
tanθ - tanθ cotθ - cotθ -tanθ tanθ - tanθ tanθ
sinθ - sinθ cosθ cosθ sinθ - sinθ - sinθ sinθ
If angle is multiple of 900
then
sin ⇔ cos;tan ⇔ cot; sec ⇔ cosec
If angle is multiple of 1800
then
sin ⇔ sin;cos ⇔ cos; tan ⇔ tan etc.

30. Trigonometric Function For Allied
Angles
Trig. ratio -θ 90o
-θ 90o
+θ 180o
-θ 180o
+θ 360o
-θ 360o
+θ
secθ secθ cosecθ - cosecθ - secθ - secθ secθ secθ
cosecθ - cosecθ secθ secθ cosecθ -cosecθ - cosecθ cosecθ
cotθ - cotθ tanθ -tanθ -cotθ cotθ - cotθ cotθ

31. Periodicity of Trigonometric
Function
Periodicity : After certain value of
x the functional values repeats
itself
Period of basic trigonometric functions
sin (360o
+θ) = sinθ ⇒ period of sinθ is 360o
or 2π
cos (360o
+θ) = cosθ ⇒ period of cosθ is 360o
or 2π
tan (180o
+θ) = tanθ ⇒ period of tanθ is 180o
or π
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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 tanA tanB
+
• + =
−
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( )
sin(A B)
Pr oof: tan A B
cos(A B)
+
− + =
+
Q
sinA cosB cos A sinB
cos A cosB sinA sinB
+
=
−

33. Trigonometric Ratio of Compound
Angle
J006
sinA cosB cos A sinB
cos A cosB sinA sinB
+
=
−
r r
Dividing N and D by cos A cosB
We get
tanA tanB
1 tanA tanB
+
=
−
Proved
( )
cotBcot A 1
cot A B
cotB cot A
−
• + =
+

34. Illustrative problem
Find the value of
(i) sin 75o
(ii) tan 105o
Solution
(i) Sin 75o
= sin (45o
+ 30o
)
= sin 45o
cos 30o
+ cos 45o
sin 30o
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 tanA tanB
−
• − =
+
Note :- by replacing B to -B in addition
formula we get difference formula
( )
cotB cot A 1
cot A B
cot A cotB
+
• − = −
−

36. Illustrative problem
If tan (θ+α) = a and tan (θ - α) = b
Prove that
a b
tan2
1 ab
−
α =
−
Solution
( ) ( ){ }tan2 tanα = θ + α − θ − α
( ) ( )
( ) ( )
tan tan
1 tan tan
θ + α − θ − α
=
+ θ + α θ − α
a b
1 ab
−
=
+

37. Some Important Deductions
• sin (A+B) sin (A-B) = sin2
A - sin2
B = cos2
B - cos2
A
• cos (A+B) cos (A-B) = cos2
A - sin2
B = cos2
B - sin2
A
( )
tanA tanB tanC tanA tanB tanC
tan A B C
1 tanA tanB tanB tanC tanC tanA
+ + −
• + + =
− − −

38. To Express acosθ + bsinθ in the
form kcosφ or λsinψ
acosθ +bsinθ
2 2
2 2 2 2
a b
a b cos sin
a b a b
 
= + θ + θ ÷
+ + 
2 2 2 2
a b
Let cos ,thensin
a b a b
α = α =
+ +
( )2 2
acos b sin a b cos cos sin sin∴ θ + θ = + θ α + θ α
( )2 2
a b cos= + θ − α
Similarly we get acosθ + bsinθ = λsinψ
2 2
k cos ,where k a b ,= φ = + φ = θ − α

39. Illustrative problem
7cosθ +24sinθ
Find the maximum and minimum
values of 7cosθ + 24sinθ
Solution
2 2
2 2 2 2
7 24
7 24 cos sin
7 24 7 24
 
= + θ + θ ÷
+ + 
7 24
25 cos sin
25 25
 
= θ + θ ÷
 
7 24
Let cos sin
25 25
α = ⇒ α =
( )7cos 24sin 25 cos cos sin sin∴ θ + θ = θ α + θ α

40. Illustrative problem
Find the maximum and minimum value of
7cosθ + 24sinθ
Solution
( )25cos 25cos where= θ − α = φ φ = θ − α
1 cos 1− ≤ φ ≤Q
25 25cos 25⇒ − ≤ φ ≤
∴ 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 cos 4x
cot 6x
sin8x sin6x sin4x
+ +
=
+ +
Solution
(cos8x cos 4x) 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
+ −
+
=
+ −
+
2cos6x cos2x cos6x
2sin6x sin2x sin6x
+
=
+
2cos6x(cos2x 1)
2sin6x(cos2x 1)
+
=
+
cot 6x= 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
π π     
θ = = = × =     
     
o
arc 2.2
radius 360 r
π
∴ θ = ⇒ =
360 2.2 360 2.2 7
r 252 cm
22
× × ×
= = =
π

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

47. Class Exercise - 3
If sinα = sinβ and cosα = cosβ, then
–
sin 0
2
α β
=(c)
–
cos 0
2
α β
=(d)
sin 0
2
α + β
=(a) cos 0
2
α + β
=(b)
Solution :-
α = βQsin sin α = βcos cosand
⇒ α β = α β =sin – sin 0 and cos – cos 0
– –
2sin cos 0 and – 2sin sin 0
2 2 2 2
α β α + β α β α + β
⇒ = =
–
sin 0
2
α β
⇒ =

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
 
= ° ° × ° 
 
[ ]
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
 
= ° + ° 
  
3
16
=
Proved.

50. Class Exercise - 5
Prove that
n n
n
cos A cosB sin A sinB
sin A – sinB cos A – cosB
A B
2 cot , if n is even
2
0, if n is odd
+ +   
+   
   
 + 
  =   


Solution :-
n n
cos A cosB sinA sinB
LHS
sinA – sinB cos A – cosB
+ +   
= +   
   
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
+ +   
   
= +   
+ +      
   
n n
A– B A – B
cot –cot
2 2
      
= +      
      

51. Class Exercise - 5
Solution :-
n n nA – B A – B
cot (–1) cot
2 2
   
= +   
   
n
n n
A – B
2cot , if n is evenA – B
cot 1 (–1) 2
2
0, if n is odd

   = + =      
Prove that
n n
n
cos A cosB sin A sinB
sin A – sinB cos A – cosB
A B
2 cot , if n is even
2
0, if n is odd
+ +   
+   
   
 + 
  =   



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 23cos x 4sinx 3 4+ = +Q
Solution :-
2 2 2 2
3 4
cos x sinx
3 4 3 4
 
+  
+ + 
3 4
5 cos x sinx
5 5
 
= + 
 
[ ]5 cosx cos sinx sin= α + α
3 4
Let cos sin
5 5
 
α = ⇒ α = 
 

53. Class Exercise - 6
The maximum value of 3 cosx + 4 sinx + 5
is
Solution :-
3 4
Let cos sin
5 5
 
α = ⇒ α = 
 
5cos(x – )= α
–1 cos(x – ) 1≤ α ≤Q
–5 5cos(x – ) 5⇒ ≤ α ≤
–5 5 5cos(x – ) 5 10⇒ + ≤ α + ≤
0 3cosx 4sinx 5 10⇒ ≤ + + ≤
∴ Maximum value of the given expression = 10.

54. Class Exercise - 7
If a and b are the solutions of a
cosθ + b sinθ = c, then show that
2 2
2 2
a – b
cos( )
a b
α+β =
+
Solution :-
We have … (i)acos bsin cθ + θ =
acos c – bsin⇒ θ = θ
2 2 2
a cos (c – bsin )⇒ θ = θ
( )2 2 2 2 2
a 1– sin c –2bcsin b sin⇒ θ = θ + θ
( )2 2 2 2 2a b sin – 2bcsin (c – a ) 0⇒ + θ θ + =
∴αβare roots of equatoin (i),

55. Class Exercise - 7
If a and b are the solutions of acosθ + bsinθ
= c, then show that α + β =
+
2 2
2 2
a – b
cos( )
a b
Solution :-
2 2
2 2
c – a
sin sin
a b
α β =
+
Hence
Again from (i),bsin c – acosθ = θ
2 2 2b sin (c – acos )⇒ θ = θ
2 2 2 2 2b (1– cos ) c a cos –2cacos⇒ θ = + θ θ
2 2 2 2 2(a b )cos – 2ac cos c – b 0⇒ + θ θ + =
∴sinα and sinβ are roots of equ. (ii).

56. Class Exercise - 7
If a and b are the solutions of acosθ + bsinθ
= c, then show that α + β =
+
2 2
2 2
a – b
cos( )
a b
Solution :- (iv)
∴α and β be the roots of equation (i),
∴cosα and cosβ are the roots of equation (iv).
2 2
2 2
c – b
cos cos
a b
α β =
+
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 sinα + β = α β α βNow

57. Class Exercise - 8
If a seca – c tana = d and b seca
+ d tana = c, then
(a) a2
+ b2
= c2
+ d2
+ cd
(c) a2
+ b2
= c2
+ d2
(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 – c tan dα α =Q
a c sin
– d
cos cos
α
⇒ =
α α
Solution :-
b sec dtan cα + α =QAgain
b dsin
c
cos cos
α
⇒ + =
α α
⇒ = α αb ccos – dsin ….. ii
a csin dcos⇒ = α + α ….. (I)
Squaring and adding
(i) and (ii), we get
2 2 2 2 2 2 2 2a b c (sin cos ) d (cos sin )+ = α + α + α + α
2cd sin cos 2cd sin cos+ α α − α α 2 2c d= +

59. Class Exercise -9
2 2A A
sin – sin –
8 2 8 2
π π   
+   
   
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
π π   
+   
   
Q
Solution :-
A A A A
sin – sin – –
8 2 8 2 8 2 8 2
   π π π π       
= + + +          
          
1
sin sin A sin A
4 2
π
= ⋅ =
2 2
sin A – sin B sin(A B)sin(A – B) = +  
Q
2 2A A
sin – sin –
8 2 8 2
π π   
+   
   
The value of

61. Class Exercise -10
If , ,
α and β lie between0 and , then value
α of tan2α is
4
cos( )
5
α + β =
5
sin( – )
13
α β =
4
π
(a) 1
(c) 0
(b)
56
33
(d)
33
56
Solution :- ∴α and βbetween 0 and ,
π
4
– – and 0
4 4 2
π π π
∴ < α β < < α + β <
Consequently, cos(α − β) and sin(α + β) are positive.
2
sin( ) 1– cos ( )α + β = α + β

62. Class Exercise -10
Solution :-
16 3
1 –
25 5
= =
2 25 12
cos( – ) 1– sin ( – ) 1–
169 13
α β = α β = =
3 5
tan( ) , tan( – )
4 12
∴ α + β = α β =
tan2 tan[( ) ( – )]α = α + β + α β
3 5
tan( ) tan( – ) 564 12
3 51– tan( )tan( – ) 33
1–
4 12
+
α + β + α β
= = =
α + β α β
×
If , ,
α and β lie between0 and , then value
α of tan2α is
4
cos( )
5
α + β =
5
sin( – )
13
α β =
4
π