This document discusses trigonometric ratios and identities. It begins by defining angles, their measurement in different systems including degrees, radians and grades. It then defines trigonometric functions including sine, cosine, tangent etc and discusses their domains, ranges and signs in different quadrants. The document also covers trigonometric identities, ratios of compound angles and periodicity of trig functions.
Here are some interesting maths quizzes. Find out the answers of these quizzes and see how genius you are....all the best!!!
Pls do visit, www. mathemagix.co.in for maths shortcuts, formulas, symbols, quizzes, facts
Here are some interesting maths quizzes. Find out the answers of these quizzes and see how genius you are....all the best!!!
Pls do visit, www. mathemagix.co.in for maths shortcuts, formulas, symbols, quizzes, facts
For a system involving two variables (x and y), each linear equation determines a line on the xy-plane. Because a solution to a linear system must satisfy all of the equations, the solution set is the intersection of these lines, and is hence either a line, a single point, or the empty set
Euclidean Geometry is another critical branch of mathematics. It weighs a lot of marks in high school math, and teachers need to teach the concept with care and inclusivity.
For a system involving two variables (x and y), each linear equation determines a line on the xy-plane. Because a solution to a linear system must satisfy all of the equations, the solution set is the intersection of these lines, and is hence either a line, a single point, or the empty set
Euclidean Geometry is another critical branch of mathematics. It weighs a lot of marks in high school math, and teachers need to teach the concept with care and inclusivity.
Had to make this dumb powerpoint for my algebra II class and I put a lot of work into it for some reason... so yeah it's just been sitting on my laptop doing nothing and I thought why not upload this to help other people? So yeah, hope you guys find it useful...
Foundations of Trigonometry: Navigating Angles and Ratios with Ease"abhishek2019pandey
Title: Unveiling the Basics of Trigonometry
I. Introduction
Definition of Trigonometry
Historical context and origins
Importance in mathematics and real-world applications
II. Fundamental Concepts
Definition of angles and their measurement
Introduction to right-angled triangles
Primary trigonometric ratios: sine, cosine, tangent
III. Trigonometric Functions
Definition of trigonometric functions
Graphs of sine, cosine, and tangent functions
Periodicity and amplitude
IV. Trigonometric Identities
Pythagorean identity
Reciprocal identities
Quotient identities
V. Solving Triangles
Use of trigonometric ratios to solve triangles
Application of the Law of Sines and Law of Cosines
Examples and practical problem-solving
VI. Applications of Trigonometry
Navigation and astronomy
Engineering and physics applications
Everyday scenarios demonstrating trigonometric principles
VII. Advanced Topics (Brief Overview)
Unit circle and radian measure
Trigonometric equations
Trigonometric functions of any angle
VIII. Interactive Examples and Demonstrations
PowerPoint slides demonstrating key concepts
Interactive activities for audience engagement
Real-life scenarios illustrating trigonometric principles
IX. Practical Tips and Tricks
Memory aids for trigonometric ratios
Problem-solving strategies
Common mistakes to avoid
X. Conclusion
Recap of key concepts
Emphasis on the practical relevance of trigonometry
Encouragement for further exploration and learning
Questions and Solutions Basic Trigonometry.pdferbisyaputra
Unlock a deep understanding of mathematics with our Module and Summary! Clear definitions, comprehensive discussions, relevant example problems, and step-by-step solutions will guide you through mathematical concepts effortlessly. Learn with a systematic approach and discover the magic in every step of your learning journey. Mathematics doesn't have to be complicated—let's make it simple and enjoyable!
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems.
This PowerPoint was created to help out graduating seniors who are taking the TAKS Mathematics Exit-Level test. It includes formulas, rules & things that they need to remember to pass the test.
Review of Trigonometry for Calculus “Trigon” =triangle +“metry”=measurement =...KyungKoh2
Review of Trigonometry for Calculus “Trigon” =triangle +“metry”=measurement =Trigonometry so Trigonometry got its name as the science of measuring triangles.
Similar to Trigonometric ratios and identities 1 (20)
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
for beginners, providing thorough training in areas such as SEO, digital communication marketing, and PPC training in Noida. After finishing the program, students receive the certifications recognised by top different universitie, setting a strong foundation for a successful career in digital marketing.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
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
J001
6. Right Angle
O
Y
X
Revolving ray describes one – quarter of a circle then
we say that measure of angle is right angle
J001
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
J001
8. System of Measurement of Angle
Measurement of Angle
Sexagesimal System
or
British System
Centesimal System
or
French System
Circular System
or
Radian Measure
J001
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
J001
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
J001
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
J002
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
J002
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 = = × = ÷
J002
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 ≠ + ∈
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 = >
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
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 = <
J004
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 = >
J004
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 = <
J004
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
J004
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 =
J004
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
J004
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
J005
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∴ − = ⇒ =
J005
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.
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
J005
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
+
• + =
−
J006
( )
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
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
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