Download Exercise -2 - teko classes bhopal

fo/u fopkjr Hkh# tu] ugha vkjEHks dke] foifr ns[k NksM+s rqjar e/;e eu dj ';keA
iq#"k flag ladYi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA
jfpr% ekuo /keZ iz.ksrk
ln~xq# Jh j.kNksM+nklth egkjkt
STUDY PACKAGE
Subject : Mathematics
Topic: Trigonometric Equation &
Properties & Solution of Triangle
Index
1. Theory
2. Short Revision
3. Exercise
4. Assertion & Reason
5. Que. from Compt. Exams
®
ENJOY
MATHEMATICS
WITH
SUHAAG SIR
Student’s Name :______________________
Class
:______________________
Roll No.
:______________________
Head Office:
243-B, III- Floor,
Near Hotel Arch Manor, Zone-I
M.P. NAGAR, Main Road, Bhopal
!:(0755) 32 00 000, 98930 58881
Free Study Package download from website : www.iitjeeiitjee.com,
www.tekoclasses.com
An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometric
equation.
Solution of Trigonometric Equation :
A solution of trigonometric equation is the value of the unknown angle that satisfies the equation.
π 3π 9π 11π
1
⇒ θ= ,
,
,
, ...........
e.g. if sinθ =
4 4
4
4
2
Thus, the trigonometric equation may have infinite number of solutions (because of their periodic nature) and
can be classified as :
(i)
Principal solution
(ii)
General solution.
2.1
Principal solutions:
The
solutions
of
a
trigonometric
equation
which
lie
in
the
interval
[0, 2π) are called Principal solutions.
1
e.g Find the Principal solutions of the equation sinx = .
2
Solution.
1
2
∵
there exists two values
π
5π
1
and
which lie in [0, 2π) and whose sine is
i.e.
6
6
2
π
1
∴
Principal solutions of the equation sinx =
are ,
6
2
General Solution :
∵
2.2
sinx =
5π
Ans.
6
The expression involving an integer 'n' which gives all solutions of a trigonometric equation is called
General solution.
General solution of some standard trigonometric equations are given below.
3.
General Solution of Some Standard Trigonometric Equations :
(i)
If sin θ = sin α
⇒ θ = n π + (−1)n α
(ii)
If cos θ = cos α
⇒ θ = 2nπ ± α
(iii)
If tan θ = tan α
⇒ θ = nπ + α
(iv)
If sin² θ = sin² α
⇒ θ = n π ± α, n ∈ Ι.
(v)
If cos² θ = cos² α
⇒ θ = n π ± α, n ∈ Ι.
(vi)
If tan² θ = tan² α
Some Important deductions :
(i)
sinθ = 0
⇒
(ii)
sinθ = 1
⇒
(iii)
sinθ = – 1
⇒
(iv)
cosθ = 0
⇒
(v)
(vi)
(vii)
cosθ = 1
cosθ = – 1
tanθ = 0
⇒
⇒
⇒
Solved Example # 1
Solve sin θ =
Solution.
3
.
2
⇒ θ = n π ± α, n ∈ Ι.
θ = nπ,
n∈Ι
π
,n ∈ Ι
2
π
θ = (4n – 1) , n ∈ Ι
2
π
θ = (2n + 1) , n ∈ Ι
2
θ = 2nπ,
n∈Ι
θ = (2n + 1)π, n ∈ Ι
θ = nπ,
n∈Ι
θ = (4n + 1)
 π π
where α ∈ − ,  ,
 2 2
where α ∈ [0, π],
 π π
where α ∈  − ,  ,
 2 2
n ∈ Ι.
n ∈ Ι.
n ∈ Ι.
[ Note: α is called the principal angle ]
98930 58881
2.
Trigonometric Equation :
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
1.
Page : 2 of 29
Trigonometric Equation
⇒
∴
3
2
π
sinθ = sin
3
Page : 3 of 29
∵
sin θ =
θ = nπ
π + (– 1)n
π
,n∈Ι
3
Ans.
Solved Example # 2
2
Solve sec 2θ
θ=–
∵
sec 2θ = –
⇒
cos2θ = –
2
3
3
2
⇒
5π
,n∈Ι
6
5π
θ = nπ
π±
,n∈Ι
12
⇒
cos2θ = cos
5π
6
Ans.
Solved Example # 3
Solve tanθ
θ=2
Solution.
∵
tanθ = 2
............(i)
Let
2 = tanα
⇒
tanθ = tanα
⇒
θ = nπ
π + α , where α = tan –1(2), n ∈ Ι
Self Practice Problems:
1.
Solve cotθ = – 1
Solve
cos3θ = –
Ans.
(1)
1
2
θ = nπ –
π
,n∈Ι
4
(2)
2nπ
2π
±
,n∈Ι
3
9
Solved Example # 4
Solve cos2θ =
Solution.
∵
cos2θ =
1
2
1
2
2
 1 

⇒
cos θ = 
 2
π
⇒
cos2θ = cos2
4
π
θ = nπ
π±
⇒
, n ∈ Ι Ans.
4
Solved Example # 5
2
Solve 4 tan2θ = 3sec2θ
Solution.
∵
4 tan2θ = 3sec2θ
.............(i)
π
For equation (i) to be defined θ ≠ (2n + 1) , n ∈ Ι
2
∵
equation (i) can be written as:
4 sin2 θ
cos θ
2
=
3
cos 2 θ
⇒
4 sin2θ = 3
⇒
 3

sin θ = 

2


2
2
∵
θ ≠ (2n + 1)
∴ cos2θ ≠ 0
π
,n∈Ι
2
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
2θ = 2nπ ±
⇒
2.
98930 58881
3
Solution.
Page : 4 of 29
π
3
π
π±
, n ∈ Ι Ans.
θ = nπ
3
sin2θ = sin2
⇒
Self Practice Problems :
1.
Solve
7cos2θ + 3 sin2θ = 4.
2.
Solve
2 sin2x + sin22x = 2
π
(1)
nπ ± , n ∈ Ι
3
Ans.
(2)
(2n + 1)
π
,n∈Ι
2
or
nπ ±
π
,n∈Ι
4
Types of Trigonometric Equations :
Type -1
Trigonometric equations which can be solved by use of factorization.
98930 58881
⇒
Solve (2sinx – cosx) (1 + cosx) = sin2x.
Solution.
∵
⇒
⇒
⇒
⇒
+ cosx) = sin2x
+ cosx) – sin2x = 0
+ cosx) – (1 – cosx) (1 + cosx) = 0
– 1) = 0
or
2sinx – 1 = 0
1
⇒
cosx = – 1
or
sinx =
2
π
⇒
x = (2n + 1)π, n ∈ Ι
or
sin x = sin
6
∴
Solution of given equation is
π
π, n ∈ Ι
π + (–1)n
(2n + 1)π
or
nπ
,n∈Ι
6
Self Practice Problems :
(2sinx – cosx) (1
(2sinx – cosx) (1
(2sinx – cosx) (1
(1 + cosx) (2sinx
1 + cosx = 0
Solve
cos3x + cos2x – 4cos2
2.
Solve
cot2θ + 3cosecθ + 3 = 0
Ans.
(1)
Type - 2
π
,n∈Ι
6
Ans.
x
=0
2
1.
(2)
⇒ x = nπ + (– 1)n
(2n + 1)π, n ∈ Ι
π
2nπ – , n ∈ Ι or
2
nπ + (–1)n + 1
π
,n∈Ι
6
Trigonometric equations which can be solved by reducing them in quadratic equations.
Solved Example # 7
Solve
Solution.
∵
⇒
⇒
2 cos2x + 4cosx = 3sin2x
∵
2cos2x + 4cosx – 3sin2x = 0
2cos2x + 4cosx – 3(1– cos2x) = 0
5cos2x + 4cosx – 3 = 0

 − 2 + 19  


 cos x −  − 2 − 19 
cos x − 

 
 = 0
5
5



 

cosx ∈ [– 1, 1] ∀ x ∈ R
∴
cosx ≠
⇒
∴
........(ii)
− 2 − 19
5
equation (ii) will be true if
cosx =
− 2 + 19
5
− 2 + 19
5
 − 2 + 19 
, n ∈ Ι
α = cos –1 

5


⇒
cosx = cosα,
where cosα =
⇒
x = 2nπ
π±α
where
Ans.
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Solved Example # 6
Ans.
4cosθ – 3secθ = tanθ
π
(1)
2nπ ± , n ∈ Ι
3
(2)
Type - 3

1 
 = 0
cos2θ – ( 2 + 1)  cos θ −
2

Page : 5 of 29
Solve
Solve
π
,n∈Ι
4
 − 1 − 17 
,n∈Ι
nπ + (– 1)n α
where α = sin–1 

8


 − 1 + 17 
, n ∈Ι
or
nπ + (–1) n β
where β = sin–1 

8


or
2nπ ±
Trigonometric equations which can be solved by transforming a sum or difference of trigonometric
ratios into their product.
Solved Example # 8
Solve cos3x + sin2x – sin4x = 0
Solution.
⇒
⇒
⇒
⇒
∴
cos3x + sin2x – sin4x = 0
cos3x – 2cos3x.sinx = 0
cos3x = 0
π
3x = (2n + 1) , n ∈ Ι
2
π
x = (2n + 1) , n ∈ Ι
6
solution of given equation is
π
(2n + 1)
,n∈Ι
or
6
⇒
⇒
or
cos3x + 2cos3x.sin(– x) = 0
cos3x (1 – 2sinx) = 0
1 – 2sinx = 0
1
sinx =
2
π
x = nπ + (–1)n , n ∈ Ι
6
or
or
nπ
π + (–1)n
π
,n∈Ι
6
Ans.
Self Practice Problems :
1.
Solve
sin7θ = sin3θ + sinθ
2.
Solve
5sinx + 6sin2x +5sin3x + sin4x = 0
3.
Solve
cosθ – sin3θ = cos2θ
nπ
(1)
,n∈Ι
3
nπ
,n∈Ι
(2)
2
2nπ
(3)
,n∈Ι
3
Ans.
or
or
or
π
nπ
±
,n∈Ι
2
12
2π
2nπ ±
,n∈Ι
3
π
2nπ – , n ∈ Ι
2
or
nπ +
π
,n∈Ι
4
Type - 4
Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their
sum or difference.
Solved Example # 9
Solve sin5x.cos3x = sin6x.cos2x
Solution.
∵
sin5x.cos3x = sin6x.cos2x
⇒
⇒
sin8x + sin2x = sin8x + sin4x
⇒
⇒
2sin2x.cos2x – sin2x = 0
⇒
⇒
sin2x = 0
or
2cos2x – 1 = 0
1
⇒
2x = nπ, n ∈ Ι or
cos2x =
2
π
nπ
⇒
x=
, n ∈ Ι or
2x = 2nπ ± , n ∈ Ι
3
2
π
⇒ x = nπ ± , n ∈ Ι
6
∴
Solution of given equation is
π
nπ
π±
,n∈Ι
or
nπ
,n∈Ι
6
2
Type - 5
2sin5x.cos3x = 2sin6x.cos2x
sin4x – sin2x = 0
sin2x (2cos2x – 1) = 0
Ans.
Trigonometric Equations of the form a sinx + b cosx = c, where a, b, c ∈ R, can be solved by dividing
both sides of the equation by
a2 + b2 .
98930 58881
2.
1.
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Self Practice Problems :
2
Solution.
sinx + cosx =
a = 1, b = 1.
∴
divide both sides of equation (i) by
1
1
sinx .
+ cosx.
=1
2
2
π
π
sinx.sin
+ cosx.cos
=1
4
4
π

cos  x −  = 1
4

⇒
⇒
⇒
⇒
∴
2
..........(i)
2 , we get
98930 58881
∵
Here
π
= 2nπ, n ∈ Ι
4
π
x = 2nπ + , n ∈ Ι
4
Solution of given equation is
π
π+
2nπ
,n∈Ι
Ans.
4
x–
Note : Trigonometric equation of the form a sinx + b cosx = c can also be solved by changing sinx and cosx
into their corresponding tangent of half the angle.
Solved Example # 11
Solve 3cosx + 4sinx = 5
Solution.
∵
∵
∴
⇒
Let
∴
⇒
⇒
⇒
⇒
⇒
⇒
⇒
3cosx + 4sinx = 5
.........(i)
x
x
2 tan
1 − tan 2
2
2
cosx =
&
sinx =
x
x
1 + tan 2
1 + tan 2
2
2
equation (i) becomes
x 


2 x 
 2 tan

 1 − tan

2 
2
+4 
=5
........(ii)
3 

2 x 

2 x 
1
tan
+
1
tan
+




2

2

x
tan
=t
2
equation (ii) becomes
 1− t 2 
2t 
 + 4 
 =5
3 
2 
 1+ t2 
 1+ t 
4t2 – 4t + 1 = 0
(2t – 1)2 = 0
1
x
t=
∵
t = tan
2
2
x
1
tan
=
2
2
x
1
tan
= tanα, where tanα =
2
2
x
= nπ + α
2
 1
π + 2α
α where α = tan–1   , n ∈ Ι
x = 2nπ
2
Self Practice Problems :
1.
Solve
3 cosx + sinx = 2
Ans.
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Solve sinx + cosx =
Page : 6 of 29
Solved Example # 10
x
=0
2
π
2nπ +
,n∈Ι
6
Solve
sinx + tan
Ans.
(1)
(2)
Page : 7 of 29
2.
x = 2nπ, n ∈ Ι
Type - 6
Trigonometric equations of the form P(sinx ± cosx, sinx cosx) = 0, where p(y, z) is a polynomial, can
be solved by using the substitution sinx ± cosx = t.
Solve sinx + cosx = 1 + sinx.cosx
Solution.
∵
Let
⇒
sinx + cosx = 1 + sinx.cosx
sinx + cosx = t
sin2x + cos 2x + 2 sinx.cosx = t 2
⇒
sinx.cosx =
Now
put
........(i)
t2 − 1
2
sinx + cosx = t
t2 − 1
2
t2 – 2t + 1 = 0
t=1
sinx + cosx = 1
and sinx.cosx =
t2 − 1
in (i), we get
2
t=1+
⇒
⇒
⇒
∵
t = sinx + cosx
.........(ii)
divide both sides of equation (ii) by 2 , we get
1
1
1
+ cosx.
=
⇒
sinx.
2
2
2
⇒
⇒
(i)
(ii)
π
π

cos  x −  = cos
4
4

π
π
x–
= 2nπ ±
4
4
if we take positive sign, we get
π
π+
x = 2nπ
,n∈Ι
Ans.
2
if we take negative sign, we get
π, n ∈ Ι
x = 2nπ
Ans.
Self Practice Problems:
1.
Solve
sin2x + 5sinx + 1 + 5cosx = 0
2.
Solve
3cosx + 3sinx + sin3x – cos3x = 0
3.
Solve
(1 – sin2x) (cosx – sinx) = 1 – 2sin2x.
π
π
(1)
nπ – , n ∈ Ι
(2)
nπ – , n ∈ Ι
4
4
π
(3)
2nπ +
,n∈Ι
or
2nπ, n ∈ Ι
or
2
Ans.
Type - 7
nπ +
π
,n∈Ι
4
Trigonometric equations which can be solved by the use of boundness of the trigonometric ratios
sinx and cosx.
Solved Example # 13
x


x


Solve sinx  cos − 2 sin x  +  1 + sin − 2 cos x  cos x = 0
4
4




Solution.
x


x


∵
sinx  cos − 2 sin x  + 1 + sin − 2 cos x  cos x = 0
.......(i)
4


4


⇒
⇒
⇒
x
x
– 2sin2x + cosx + sin .cosx – 2cos 2x = 0
4
4
x
x


 sin x. cos + sin . cos x  – 2 (sin2x + cos2x) + cosx = 0
4
4


sinx.cos
sin
5x
+ cosx = 2
4
........(ii)
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
98930 58881
Solved Example # 12
Page : 8 of 29
and
cosx = 1
x = 2mπ, m ∈ Ι
and
x = 2mπ, m ∈ Ι
........(iv)
98930 58881
and
– 4, p ∈ Ι
∴
general solution of given equation can be obtained by substituting either m = 4p – 3 in
equation (iv) or n = 5p – 4 in equation (iii)
∴
general solution of equation (i) is
π, p ∈ Ι
(8p – 6)π
Ans.
Self Practice Problems :
1.
Solve
sin3x + cos2x = – 2
2.
Solve
3 sin 5 x − cos 2 x − 3 = 1 – sinx
π
(2)
(1)
(4p – 3) , p ∈ Ι
2
Ans.
2mπ +
π
,m∈Ι
2
Exercise -1
Part : (A) Only one correct option
1.
The solution set of the equation 4sinθ.cosθ – 2cosθ – 2 3 sinθ + 3 = 0 in the interval (0, 2π) is
 3π 7π 

(A)  ,
4 4
2.
2π
,n∈Ι
3
(C) nπ or m π ±
5.
π
where n, m ∈ Ι
3
If 20 sin2 θ + 21 cos θ − 24 = 0 &
(A) 3
4.
π 5π 
 3π

(C)  , π, ,
4
3
3 

 π 5 π 11π 
,

(D)  ,
6 6 6 
All solutions of the equation, 2 sinθ + tanθ = 0 are obtained by taking all integral values of m and n in:
(A) 2nπ +
3.
 π 5π 

(B)  ,
3 3 
(B)
(B) nπ or 2m π ±
2π
where n, m ∈ Ι
3
(D) nπ or 2m π ±
π
where n, m ∈ Ι
3
7π
θ
< θ < 2π then the values of cot is:
4
2
15
3
(C) −
15
3
The general solution of sinx + sin5x = sin2x + sin4x is:
(A) 2 nπ ; n ∈ Ι
(B) nπ ; n ∈ Ι
(C) nπ/3 ; n ∈ Ι
A triangle ABC is such that sin(2A + B) =
(D) − 3
(D) 2 nπ/3 ; n ∈ Ι
1
. If A, B, C are in A.P. then the angle A, B, C are
2
respectively.
(A)
5π π π
, ,
12 4 3
(B)
π π 5π
, ,
4 3 12
(C)
π π 5π
, ,
3 4 12
(D)
π 5π π
,
,
3 12 4
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Now equation (ii) will be true if
5x
sin
=1
4
π
5x
= 2nπ +
,n∈Ι
⇒
2
4
(8n + 2)π
⇒
x =
,n∈Ι
........(iii)
5
Now to find general solution of equation (i)
(8n + 2)π
= 2mπ
5
⇒
8n + 2 = 10m
5m − 1
⇒
n=
4
if
m=1
then
n=1
if
m=5
then
n=6
.........
.........
.........
.........
.........
.........
if
m = 4p – 3, p ∈ Ι
then
n = 5p
8.
(D) 7
If sin θ + 7 cos θ = 5, then tan (θ/2) is a root of the equation
(B) 6x2 − x − 1 = 0
(C) 6x 2 + x + 1 = 0
(A) x2 − 6x + 1 = 0
(D) x2 − x + 6 = 0
sin 3 θ − cos 3 θ
cos θ
−
− 2 tan θ cot θ = − 1 if:
sin θ − cos θ
1 + cot 2 θ


(A) θ ∈  0 ,
π

2
π 
, π
2 
(B) θ ∈ 


(C) θ ∈  π ,
 3π

, 2 π
 2

3π 

2
(D) θ ∈ 
9.
The number of integral values of a for which the equation cos 2x + a sin x = 2a − 7 possesses a solution
is
(A) 2
(B) 3
(C) 4
(D) 5
10.
The principal solution set of the equation, 2 cos x = 2 + 2 sin 2 x is
 π 13 π 
(A)  ,

8 8 
11.
 π 13 π 

(B)  ,
4 8 
 π 13 π 

(C)  ,
 4 10 
 π 13 π 

(D)  ,
 8 10 
The number of all possible triplets (a1, a2, a 3) such that : a 1 + a2 cos 2x + a 3 sin2x = 0 for all x is
(A) 0
(B) 1
(C) 2
(D) infinite
12.
 nπ 
If 2tan2x – 5 secx – 1 = 0 has 7 different roots in 0,
, n ∈ N, then greatest value of n is
2 

(A) 8
(B) 10
(C) 13
(D) 15
13.
The solution of |cosx| = cosx – 2sinx is
(A) x = nπ, n ∈ Ι
(C) x = nπ + (–1)n
(B) x = nπ +
π
,n∈Ι
4
π
,n∈Ι
4
(D) (2n + 1)π +
π
,n∈Ι
4
14.
The arithmetic mean of the roots of the equation 4cos3x – 4cos2x – cos(π + x) – 1 = 0 in the interval
[0, 315] is equal to
(A) 49π
(B) 50π
(C) 51π
(D) 100π
15.
Number of solutions of the equation cos 6x + tan2 x + cos 6x . tan2 x = 1 in the interval [0, 2π] is :
(A) 4
(B) 5
(C) 6
(D) 7
Part : (B) May have more than one options correct
16.
sinx − cos 2x − 1 assumes the least value for the set of values of x given by:
(B) x = nπ + (−1) n (π/6) , n ∈ Ι
(A) x = nπ + (−1)n+1 (π/6) , n ∈ Ι
n
(C) x = nπ + (−1) (π/3), n ∈ Ι
(D) x = nπ − (−1)n (π/6) , n ∈ Ι
17.
cos4x cos8x − cos5x cos9x = 0 if
(A) cos12x = cos 14 x
(C) sinx = 0
18.
The equation 2sin
(A) sin2x = 1
19.
20.
Page : 9 of 29
(C) 5
(B) sin13 x = 0
(D) cosx = 0
x
x
. cos2x + sin 2x = 2 sin . sin2x + cos2x has a root for which
2
2
1
1
(B) sin2x = – 1
(C) cosx =
(D) cos2x = –
2
2
sin 2x + 2 sin x cos x − 3cos2x = 0 if
(A) tan x = 3
(C) x = nπ + π/4, n ∈ Ι
(B) tanx = − 1
(D) x = nπ + tan −1 (−3), n ∈ Ι
sin 2x − cos 2x = 2 − sin 2x if
(A) x = nπ/2, n ∈ Ι
(C) x = (2n + 1) π/2, n ∈ Ι
(B) x = nπ − π/2, n ∈ Ι
(D) x = nπ + (−1) n sin−1 (2/3), n ∈ Ι
98930 58881
7.
The maximum value of 3sinx + 4cosx is
(A) 3
(B) 4
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
6.
Solve
cotθ = tan8θ
x
x
cot   – cosec   = cotx
2
 
2
2.
Solve
3.
Solve

1 
 cotθ + 1 = 0.
cot 2θ +  3 +

3

4.
Solve
cos2θ + 3 cosθ = 0.
5.
Solve the equation: sin 6x = sin 4x − sin 2x .
6.
Solve: cos θ + sin θ = cos 2 θ + sin 2 θ .
7.
Solve
4 sin x . sin 2x . sin 4x = sin 3x .
8.
Solve
sin 2nθ – sin2(n – 1)θ = sin2θ, where n is constant and n ≠ 0, 1
9.
Solve
tanθ + tan2θ +
10.
Solve: sin 3 x cos 3 x + cos3 x sin 3 x + 0.375 = 0
11.
sin 3 x − cos 3 x cos x
2
2 =
Solve the equation,
.
3
2 + sin x
12.
Solve the equation: sin 5x = 16 sin5 x .
13.
If tan θ + sin φ =
14.
Solve for x, the equation
15.
Find the general solution of sec 4 θ − sec 2 θ = 2 .
16.
Solve the equation
17.
Solve for x: 2 sin  3 x +
18.
Solve the equation for 0 ≤ θ ≤ 2 π; sin 2θ + 3 cos2θ
19.
Solve: tan 2 x . tan 2 3 x . tan 4 x = tan2 x − tan2 3 x + tan 4 x .
20.
Find the values of x, between 0 & 2 π, satisfying the equation; cos 3x + cos 2x = sin
21.
Solve: cos
22.
Solve the equation, sin2 4 x + cos2 x = 2 sin 4 x cos4 x .
3.
7
3
& tan² θ + cos² φ =
then find the general value of θ & φ .
4
2


13 − 18 tan x = 6 tan x − 3, where − 2 π < x < 2 π .
3
sin x − cos x = cos² x .
2
π
 =
4
1 + 8 sin 2 x . cos 2 2 x .
(
)
2

π
− 5 = cos  − 2θ  .
6


2x
cos 6 x = − 1 .
3
3x
x
+ sin .
2
2
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
3 tanθ tan2θ =
98930 58881
1.
Page : 10 of 29
Exercise -2
Page : 11 of 29
ANSWER
EXERCISE 1
3. D
4. C
5. B
6. C
7. B
8.
11. D
12. D
13. D
14. C
15. D
16. AD 17. ABC
B
9.
18.
ABCD
D
10.
A
19.
CD
98930 58881
2. B
20. BC
EXERCISE 2
1.
1 π

n +  , n ∈ Ι
2 9

3. θ = nπ –
5.
π
π
, n ∈ Ι or nπ – , n ∈ Ι
3
6
π
nπ
, n ∈ Ι or n π ± , n ∈ Ι
6
4
7. x = n π, n ∈ Ι or
9.
13. θ = n π +
π
,n∈Ι
2
π
π
, φ = n π + (−1) n , n ∈ I
4
6
2nπ
π
π
±
or 2nπ ± , n ∈ Ι
5
10
2
17. (24" + 1)
19.
π
nπ
± , n∈Ι
9
3
1 π

n + 
,n∈Ι
3 3

11. x = (4 n + 1)
15.
2. x = 4nπ ±
21. φ
 17 − 3 
, n ∈ Ι
4. 2nπ ± α where α = cos–1 

4


2nπ π
+ , n∈Ι
3
6
6. 2 n π, n ∈ Ι or
8. mπ, m ∈ Ι or
10. x =
1 π

mπ
, m ∈ Ι or  m + 
,m∈Ι
2
 n

n −1
nπ
π
+ (− 1)n + 1
, n∈Ι
4
24
12. x = n π ; x = n π ±
π
,n∈Ι
6
14. α − 2 π; α − π, α, α + π, where tan α =
16. x = (2 n + 1)π, , n ∈ Ι
π
π
, " ∈ Ι or x = (24k – 7)
, k∈Ι
12
12
(2 n + 1) π
, k π, where n, k ∈ Ι
4
2π
,n∈Ι
3
20.
18.
θ=
7 π 19 π
,
12
12
π 5π
9 π 13 π
,
,π,
,
7 7
7
7
22. x = (2 n + 1)
π
, n∈I
2
or
2
3
2nπ ±
π
, n∈ Ι
3
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
1. D
1. Sine Rule:
In any triangle ABC, the sines of the angles are proportional to the opposite sides i.e.
a
b
c
=
=
.
sin A sin B sin C
=
98930 58881
Solution.
k(sin A + sin B)
k sin C
C
 A −B
cos 

2
 2 
=
C
C
sin cos
2
2
= R.H.S.
Hence L.H.S. = R.H.S.
Proved
In any ∆ABC, prove that
(b2 – c2) cot A + (c2 – a 2) cot B + (a2 – b 2) cot C = 0
∵
We have to prove that
(b2 – c2) cot A + (c2 – a 2) cot B + (a2 – b 2) cot C = 0
∵
from sine rule, we know that
a = k sinA, b = k sinB and c = k sinC
∴
(b2 – c2) cot A = k2 (sin2B – sin2C) cot A
∵
sin 2B – sin2C = sin (B + C) sin (B – C)
∴
(b2 – c2) cot A = k2 sin (B + C) sin (B – C) cotA
cos A
∴
(b2 – c2) cot A = k2 sin A sin (B – C)
sin A
= – k2 sin (B – C) cos (B + C)
cos
Example :
Solution.
k2
[2sin (B – C) cos (B + C)]
2
k2
⇒
(b2 – c2) cot A = –
[sin 2B – sin 2C]
2
k2
Similarly
(c2 – a2) cot B = –
[sin 2C – sin 2A]
2
k2
and
(a2 – b2) cot C = –
[sin 2A – sin 2B]
2
adding equations (i), (ii) and (iii), we get
(b2 – c2) cot A + (c2 – a 2) cot B + (a2 – b 2) cot C = 0
Self Practice Problems
In any ∆ABC, prove that

A
A
1.
a sin  + B  = (b + c) sin   .

2
2
 A +B
 A −B
sin 
 cos 

 2 
 2 
=
C
C
sin cos
2
2
 A −B
cos 

 2 
=
C
sin
2
∵
B+C=π–A
∵
cosA = – cos(B + C)
=–
2.
a 2 sin(B − C)
b 2 sin(C − A )
c 2 sin( A − B)
+
+
=0
sin B + sin C
sin C + sin A
sin A + sin B
2. Cosine Formula:
(i) cos A =
b 2 + c2 − a 2
2bc
3.
..........(i)
..........(ii)
..........(iii)
Hence Proved
A
B
tan + tan
c
2
2 .
=
A
B
a −b
tan − tan
2
2
or a² = b² + c² − 2bc cos A = b 2 + c2 + 2bc cos (B + C)
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Example :
 A −B
cos 

a+b
 2 
In any ∆ABC, prove that
=
.
C
c
sin
2
 A −B
cos 

a+b
 2 
∵
We have to prove
=
.
C
c
sin
2
∵
From sine rule, we know that
a
b
c
=
=
= k (let)
sin A
sin B
sin C
⇒
a = k sinA, b = k sinB and c = k sinC
a+b
∵
L.H.S. =
c
Page : 12 of 29
Properties & Solution of Triangle
∵
Page : 13 of 29
Solution.
cosC
∴
=
a2 + b2 − c 2
2ab
&
cos B =
98930 58881
*Example :
Solution.
64 + 49 − 169
b2 + c 2 − a2
=
2.8.7
2bc
2π
1
⇒
cosA = –
⇒
A=
3
2
2π
3
∴
sinA = sin
=
Ans.
3
2
In a ∆ABC, prove that a(b cos C – c cos B) = b2 – c2
∵
We have to prove a (b cosC – c cosB) = b2 – c2.
∵
from cosine rule we know that
cosA =
a2 + c 2 − b2
2ac
  a 2 + b 2 − c 2 
 a2 + c 2 − b2



b
−
c
L.H.S. = a  


2ab
2ac
 


a 2 + b2 − c 2
(a 2 + c 2 − b 2 )
–
2
2
= (b2 – c2)
Hence L.H.S. = R.H.S.
Proved




=
Example :
Solution.
= R.H.S.
c a
 a b 
If in a ∆ABC, ∠A = 60° then find the value of 1 + +  1 + −  .
c c  b b

∵
∠A = 60°
c a
 a b 
c +a+b b+c −a
 

1 + +  1 + −  = 
∵
c c  b b
c
b
 



=
(b + c )2 − a 2
bc
=
(b 2 + c 2 − a 2 ) + 2bc
bc
=
b2 + c 2 − a2
+2
bc
 b2 + c 2 − a 2 


=2 
 +2
2
bc


= 2cosA + 2
∵ ∠A = 60°
⇒
cos A =
1
2
 a b  c a
1 + +  1 + −  = 3 Ans.
 c c  b b
Self Practice Problems :
∴
a 2 + ab + b 2 , then prove that the greatest angle is 120°.
A
a(cosB + cosC) = 2(b + c) sin2
.
2
1.
The sides of a triangle ABC are a, b,
2.
In a triangle ABC prove that
3.
Projection Formula:
(i) a = b cosC + c cosB
(ii) b = c cosA + a cosC
(iii) c = a cosB + b cosA
Example :
In a triangle ABC prove that a(b cosC – c cosB) = b2 – c2
Solution.
∵
L.H.S. = a (b cosC – c cosB)
= b (a cosC) – c (a cosB)
............(i)
∵
From projection rule, we know that
b = a cosC + c cosA
⇒
a cosC = b – c cosA
&
c = a cosB + b cosA
⇒
a cosB = c – b cosA
Put values of a cosC and a cosB in equation (i), we get
L.H.S. = b (b – ccos A) – c(c – b cos A)
= b2 – bc cos A – c2 + bc cos A
= b2 – c2
= R.H.S.
Hence L.H.S. = R.H.S.
Proved
Note: We have also proved a (b cosC – ccosB) = b2 – c2 by using cosine – rule in solved *Example.
Example :
In a ∆ABC prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c.
Solution.
∵
L.H.S. =
=
=
=
=
Hence L.H.S. =
(b + c) cos A (c + a) cos B + (a + B) cos C
b cos A + c cos A + c cos B + a cos B + a cos C + b cos C
(b cos A + a cos B) + (c cos A + a cos C) + (c cos B + b cos C)
a+b+c
R.H.S.
R.H.S.
Proved
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
Example :
c2 + a 2 − b 2
a 2 + b 2 − c2
(iii) cos C =
2 ca
2a b
In a triangle ABC if a = 13, b = 8 and c = 7, then find sin A.
(ii) cos B =
2.
cos B
c − b cos A
=
.
cos C
b − c cos A
3.
cos A
cos B
cos C
a2 + b2 + c 2
+
+
=
.
c cos B + b cos C
a cos C + c cos A
a cos B + b cos A
2abc
4. Napier’s Analogy - tangent rule:
c −a
B
B−C
A
C−A
b−c
=
cot
(ii) tan
=
cot
c +a
2
2
2
2
b+c
A−B a−b
C
(iii) tan
=
cot
+
a
b
2
2
Example :
Find the unknown elements of the ∆ABC in which a = 3 + 1, b = 3 – 1, C = 60°.
(i) tan
Solution.
∵
a=
∵
3 + 1, b = 3 – 1, C = 60°
A + B + C = 180°
∴
A + B = 120°
∵
From law of tangent, we know that
.......(i)
a−b
C
 A −B 
 =
cot
tan 
a
+
b
2
 2 
=
=
( 3 + 1) − ( 3 − 1)
( 3 + 1) + ( 3 − 1)
2
2 3
cot 30°
cot 30°
 A −B 
 =1
tan 
 2 
π
A −B
∴
=
= 45°
4
2
⇒
A – B = 90°
From equation (i) and (ii), we get
A = 105°
and
B = 15°
Now,
⇒
∵
From sine-rule, we know that
∴
c=
.......(ii)
a
b
c
=
=
sin A
sin B
sin C
a sin C
( 3 + 1) sin 60°
=
sin A
sin105°
3
2
3 +1
( 3 + 1)
=
∵
sin105° =
3 +1
2 2
2 2
⇒
c=
∴
c=
Self Practice Problem
1.
6 , A = 105°, B = 15°
In a ∆ABC if b = 3, c = 5 and cos (B – C) =
Ans.
2.
6
Ans.
7
A
, then find the value of tan
.
25
2
1
3
A
B
C
B −C
C−A
 A −B 
If in a ∆ABC, we define x = tan 
 tan
, y = tan 
 tan
and z = tan 
 tan
2
2
2
2
2
2






then show that x + y + z = – xyz.
98930 58881
In a ∆ABC, prove that
B

2 C
+ c cos 2  = a + b + c.
2  b cos
2
2

TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
1.
Page : 14 of 29
Self Practice Problems
(i)
sin
(s − c) (s − a )
(s − b) (s − c)
A
B
C
=
; sin
=
; sin
=
ca
b
c
2
2
2
(ii)
cos
s (s − b )
s (s − a )
A
B
C
=
; cos
=
; cos
=
ca
b
c
2
2
2
(iii)
tan
A
=
2
(iv)
sin A =
(s − a ) (s − b)
ab
s (s − c)
ab
(s − b) (s − c)
∆
a+b+c
=
where s =
is semi perimetre of triangle.
s (s − a )
s (s − a )
2
2∆
bc
98930 58881
s(s − a )(s − b)(s − c) =
1
1
1
ab sin C = bc sin A = ca sin B = s (s − a ) (s − b) (s − c)
2
2
2
Example :
In a ∆ABC if a, b, c are in A.P. then find the value of tan
Solution.
∵
tan
∴
tan
∴
∵
⇒
∵
∴
∴
⇒
∆
A
=
− a)
s
(
s
2
and tan
A
C
. tan
.
2
2
∆
C
=
− c)
s
(
s
2
∆2
A
C
. tan
= 2
s (s − a)(s − c )
2
2
s−b
b
A
C
tan
. tan
=
=1–
s
s
2
2
it is given that a, b, c are in A.P.
2b = a + c
a+b+c
3b
s=
=
2
2
b
2
=
put in equation (i)
s
3
2
A
C
tan
. tan
=1–
3
2
2
1
A
C
tan
. tan
=
Ans.
3
2
2
∵
∆2 = s (s – a) (s – b) (s – c)
........(i)
Example :
In a ∆ABC if b sinC(b cosC + c cosB) = 42, then find the area of the ∆ABC.
Solution.
∵
∵
∵
∴
b sinC (b cosC + c cosB) = 42
From projection rule, we know that
a = b cosC + c cosB put in (i), we get
ab sinC = 42
1
∆=
ab sinC
2
∆ = 21 sq. unit
Ans.
........(i) given
........(ii)
Example :
C
A
B

In any ∆ABC prove that (a + b + c)  tan + tan  = 2c cot .
2
2
2

Solution.
∵
A
B

L.H.S. = (a + b + c)  tan + tan 
2
2

∵
tan
∴
A
2
=
(s − b)(s − c )
s(s − a)

L.H.S. = (a + b + c) 

s−c 

= 2s
s


= 2 s( s − c ) 

and tan
B
=
2
(s − a)(s − c )
s(s − b)
(s − b)(s − c )
(s − a)(s − c ) 
+

s(s − a)
s(s − b) 
s−b
s−a
+

s−a
s − b 
s−b + s−a 

(s − a)(s − b) 
∵
2s= a + b + c
∴
2s – b – a = c
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
2
bc
6. Area of Triangle (∆)
∆=
Page : 15 of 29
5. Trigonometric Functions of Half Angles:
s(s − c )
(s − a)(s − b)
= 2c cot
= R.H.S.
Hence L.H.S. = R.H.S.
∵
cot
C
=
2
s(s − c )
(s − a)(s − b)
C
2
Proved
98930 58881
7. m - n Rule:
(m + n) cot θ = m cot α − n cot β
= n cot B − m cot C
Example :
If the median AD of a triangle ABC is perpendicular to AB, prove that tan A + 2tan B = 0.
Solution.
From the figure, we see that θ = 90° + B (as θ is external angle of ∆ABD)
Now if we apply m-n rule in ∆ABC, we get
(1 + 1) cot (90 + B) = 1. cot 90° – 1.cot (A – 90°)
⇒
– 2 tan B = cot (90° – A)
⇒
– 2 tan B = tan A
⇒
tan A + 2 tan B = 0
Hence proved.
Example :
The base of a triangle is divided into three equal parts. If t 1, t2, t3 be the tangents of the angles
subtended by these parts at the opposite vertex, prove that

1 1
1
4 1 + 2  =  + 
 t1 t 2 
 t2 
Solution.
1
1
 + .
t

 2 t3 
Let point D and E divides the base BC into three equal parts i.e. BD = DE = DC = d (Let) and
let α, β and γ be the angles subtended by BD, DE and EC respectively at their opposite vertex.
⇒
t1 = tanα, t2 = tanβ and t3 = tanγ
Now in ∆ABC
∵
BE : EC = 2d : d = 2 : 1
∴
from m-n rule, we get
(2 + 1) cotθ = 2 cot (α + β) – cotγ
⇒
3cotθ = 2 cot (α + β) – cotγ
.........(i)
again
∵
in ∆ADC
∵
DE : EC = x : x = 1 : 1
∴
if we apply m-n rule in ∆ADC, we get
(1 + 1) cotθ = 1. cotβ – 1 cotγ
2cotθ = cotβ – cotγ
.........(ii)
from (i) and (ii), we get
2 cot(α + β) − cot γ
3 cot θ
=
cot β − cot γ
2 cot θ
⇒
3cotβ – 3cotγ = 4cot (α + β) – 2 cotγ
⇒
3cotβ – cotγ = 4 cot (α + β)
 cot α. cot β − 1

⇒
3cotβ – cotγ = 4 
 cot β + cot α 
⇒
3cot2β + 3cotα cotβ – cotβ cotγ – cotα cotγ = 4 cotα cotβ – 4
⇒
4 + 3cot2β = cotα cotβ + cotβ cotγ + cotα cotγ
⇒
4 + 4cot 2β = cotα cotβ + cotα cotγ + cotβ cotγ + cot2β
⇒
4(1 + cot 2β) = (cotα + cotβ) (cotβ + cotγ)

 1
1   1
1 
1 


 

+
+
⇒
4 1 +
2  =  tan α
tan β   tan β tan γ 
tan β 


⇒
Page : 16 of 29
= 2c


c

s(s − c ) 
 (s − a)(s − b) 

1 1 1
1
1
4 1 + 2  =  +   + 
 t1 t 2   t 2 t 3 
 t2 
Hence proved
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
=2
1
In a ∆ABC, the median to the side BC is of length
11 − 6 3
30° and 45°. Prove that the side BC is of length 2 units.
and it divides angle A into the angles of
8. Radius of Circumcirlce :
a
b
c
a bc
=
=
=
2 sinA 2 sinB 2 sinC
4∆
s
R
Example :
In a ∆ABC prove that sinA + sinB + sinC =
Solution.
In a ∆ABC, we know that
a
b
c
=
=
= 2R
sin A
sin B
sin C
a
b
c
∴
sin A =
, sinB =
and sinC =
.
2R
2R
2R
a+b+c
∴
sinA + sinB + sinC =
∵
a + b + c = 2s
2R
2s
s
=
⇒
sinA + sinB + sinC =
.
2R
R
In a ∆ABC if a = 13 cm, b = 14 cm and c = 15 cm, then find its circumradius.
abc
.......(i)
∵
R=
4∆
∵
∆ = s( s − a)(s − b)(s − c )
Example :
Solution.
a+b+c
= 21 cm
2
∴
∆ = 21.8.7.6 = 7 2.4 2.3 2
⇒
∆ = 84 cm2
13.14.15
65
∴
R=
=
cm
4.84
8
65
∴
R=
cm.
8
A
B
C
In a ∆ABC prove that s = 4R cos . cos . cos .
2
2
2
In a ∆ABC,
∵
Example :
Solution.
s(s − a)
s(s − b)
B
C
, cos
=
and cos
=
bc
ca
2
2
A
B
C
∵
R.H.S. = 4R cos . cos . cos .
2
2
2
s
(
s
−
a
)(
s
−
b
)(
s − c)
abc
.s
∵
=
2
(abc )
∆
= s
= L.H.S.
Hence R.H.L = L.H.S. proved
1
1
1
1
4R
In a ∆ABC, prove that
+
+
–
=
.
s−a
s−b
s−c
s
∆
4R
1
1
1
1
+
+
–
=
∆
s−a
s−b
s−c
s
1 
1
 1
 1
+
− 
 + 
∵
L.H.S. = 
s
−
a
s
−
b
s
−
c
s



∵
Example :
Solution.
s=
cos
A
=
2
2s − a − b
(s − s + c )
+
(s − a)(s − b)
s(s − c )
c
c
=
+
(s − a)(s − b)
s(s − c )
=
∵
s(s − c )
abc
and R =
ab
4∆
∆=
s( s − a )(s − b )(s − c )
2s = a + b + c
 2s 2 − s(a + b + c ) + ab 
 s(s − c ) + (s − a)(s − b) 

=c 
 =c 
∆2

 s(s − a)(s − b)(s − c ) 

∴
 2s 2 − s(2s) + ab 
abc
4R∆
4R
L.H.S. = c 
=
 = 2 =
2
2
∆
∆
∆
∆


∵
⇒
abc
4∆
abc = 4R∆
R=
98930 58881
R=
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
1.
Page : 17 of 29
Self Practice Problems :
Page : 18 of 29
Self Practice Problems :
4R
∆
In a ∆ABC, prove the followings :
1.
a cot A + b cotB + cos C = 2(R + r).
2.
s  s  s 
r
4  − 1  − 1  − 1 =
.

a  b  c
R
3.
If α, β, γ are the distances of the vertices of a triangle from the corresponding points of contact with the
αβ y
incircle, then prove that
= r2
α+β+ y
9. Radius of The Incircle :
∆
s
a sin B2 sin C2
(iii) r =
cos A2
A
B
C
= (s − b) tan
= (s − c) tan
2
2
2
A
B
C
(iv) r = 4R sin
sin
sin
2
2
2
(ii) r = (s − a) tan
(i) r =
& so on
10. Radius of The Ex- Circles :
A
B
C
∆ ;
∆ ;
∆
r2 =
r3 =
(ii) r1 = s tan ; r2 = s tan ; r3 = s tan
2
2
2
s−a
s−b
s−c
B
C
a cos 2 cos 2
A
B
C
(iii) r 1 =
& so on
(iv) r1 = 4 R sin . cos . cos
2
2
2
cos A2
Example :
In a ∆ABC, prove that r1 + r2 + r3 – r = 4R = 2a cosecA
(i) r 1 =
Solution.
∵
L.H.S
= r 1 + r2 + r3 – r
∆
∆
∆
∆
+
+
–
=
s−a
s−b
s−c
s
1 
1
 1
 1
+
− 
 +∆ 
=∆ 
s
−
a
s
−
b
s
−
c
s



 s − b + s − a   s − s + c 
= ∆  (s − a)(s − b)  +  s(s − c ) 

 


c
c 
+
= ∆ 

(
s
−
a
)(
s
−
b
)
s
(
s
− c) 

 s(s − c ) + (s − a)(s − b) 
= c∆ 

 s(s − a)( s − b)(s − c ) 
 2s 2 − s(a + b + c ) + ab 

= c∆ 
∆2


abc
=
∆
∵
= 4R = 2acosecA
∵
a + b + c = 2s
∵
R=
abc
4∆
a
= 2R = acosecA
sin A
Example :
= R.H.S.
Hence L.H.S. = R.H.S.
proved
If the area of a ∆ABC is 96 sq. unit and the radius of the escribed circles are respectively
8, 12 and 24. Find the perimeter of ∆ABC.
Solution.
∵
∵
∵
∵
∴
∴
∆ = 96 sq. unit
r1 = 8, r2 = 12 and r3 = 24
∆
r1 =
⇒
s – a = 12
s−a
∆
r2 =
⇒
s–b=8
s−b
∆
r3 =
⇒
s–c=4
s−c
adding equations (i), (ii) & (iii), we get
3s – (a + b + c) = 24
s = 24
perimeter of ∆ABC = 2s = 48 unit.
.........(i)
.........(ii)
.........(iii)
98930 58881
L.H.S. =
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
∴
Page : 19 of 29
Self Practice Problems
In a ∆ABC prove that
1.
r 1r 2 + r 2r 3 + r 3 r1 = s2
2.
rr1 + rr2 + rr3 = ab + bc + ca – s2
3.
If A, A1, A2 and A3 are the areas of the inscribed and escribed circles respectively of a ∆ABC, then prove
1
1
1
1
that
=
+
+
.
A
A1
A2
A3
98930 58881
c
r1 − r
r2 − r
+
= r .
a
b
3
11. Length of Angle Bisectors, Medians & Altitudes :
(i) Length of an angle bisector from the angle A = βa =
2 bc cos A
2
b+c
;
1
2 b2 + 2 c2 − a 2
2
2∆
&
(iii) Length of altitude from the angle A = Aa =
a
3
2
2
2
NOTE : ma + m b + m c =
(a2 + b2 + c2)
4
(ii) Length of median from the angle A = ma =
Example :
AD is a median of the ∆ABC. If AE and AF are medians of the triangles ABD and ADC
respectively, and AD = m 1, AE = m2 , AF = m3 , then prove that m 22 + m32 – 2m12 =
Solution.
In ∆ABC
1
AD 2 =
(2b 2 + 2c 2 – a2) = m12
4
1
a2
∵
In ∆ABD, AE2 = m22 =
(2c2 + 2AD2 –
)
4
4
2
1  2AD 2 + 2b 2 − a 
Similarly in ∆ADC, AF2 = m 32 =
4 
4 
by adding equations (ii) and (iii), we get
∵
∵
a2
.
8
.........(i)
.........(ii)
........(iii)
2 

 4AD 2 + 2b 2 + 2c 2 − a 

2 

2
1  2b 2 + 2c 2 − a 
2
= AD +

2 
4 
2
1  2b 2 + 2c 2 − a 2 + a 
2
= AD +

2 
4 
m 22 + m 32 =
1
4
1
a2
(2b2 + 2c2 – a2) +
4
8
2
a
= AD2 + AD2 +
8
2
a
= 2AD2 +
∵
8
a2
= 2m 12 +
8
= AD2 +
∴
m 22 + m32 – 2m12 =
a2
8
AD 2 = m12
Hence Proved
Self Practice Problem :
3.
In a ∆ABC a = 5, b = 4, c = 3. ‘G’ is the centroid of triangle, then find circumradius of ∆GAB.
5
Ans.
13
12
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
4.
:
OA = R & Oa = R cos A
(ii)
Incentre (I)
:
IA = r cosec
(iii)
Excentre (I1)
:
(iv)
Orthocentre (H)
:
HA = 2R cos A & Ha = 2R cos B cos C
(v)
Centroid (G)
:
GA =
Example :
Solution.
A
& Ia = r
2
A
I1 A = r1 cosec
& I1a = r1
2
2∆
1
2b2 +2c2 −a 2 & G a =
3a
3
If x, y and z are respectively the distances of the vertices of the ∆ABC from its orthocentre,
then prove that
abc
a
c
b
(i)
+
+
=
(ii)
x y + z = 2(R + r)
xyz
x
z
y
∵
x = 2R cosA, y = 2R cosB, z = 2R cosC
and
and
a = 2R sinA, b = 2R sinB, c = 2R sinC
a
c
b
∴
+
+
= tanA + tan B + tan C
.........(i)
x
z
y
&
∵
∴
∵
∵
∴
∴
abc
........(ii)
xyz = tanA. tanB. tanC
Σ tanA = Π tanA
We know that in a ∆ABC
From equations (i) and (ii), we get
abc
a
c
b
+
+
=
xyz
x
z
y
x + y + z = 2R (cosA + cosB + cosC)
A
B
C
in a ∆ABC
cosA + cosB + cosC = 1 + 4sin sin sin
2
2
2
A
B
C

x + y + z = 2R 1 + 4 sin . sin . sin 
2
2
2

A
B
C

= 2  R + 4R sin . sin . sin 
2
2
2

x + y + z = 2(R + r)
∵
r = 4R sin
B
C
A
sin
sin
2
2
2
Self Practice Problems
A
B
C
tan
tan .
2
2
2
1.
If Ι be the incentre of ∆ABC, then prove that ΙA . ΙB . ΙC = abc tan
2.
If x, y, z are respectively be the perpendiculars from the circumcentre to the sides of ∆ABC, then prove
abc
a
c
b
+
+
=
.
that
4 xyz
x
z
y
13. Orthocentre and Pedal Triangle:
The triangle KLM which is formed by joining the feet of the altitudes is called the Pedal Triangle.
(i) Its angles are π − 2A, π − 2B and π − 2C.
(ii) Its sides are a cosA = R sin 2A,
b cosB = R sin 2B and
c cosC = R sin 2C
(iii) Circumradii of the triangles PBC, PCA, PAB and ABC are equal.
14. Excentral Triangle:
The triangle formed by joining the three excentres Ι1, Ι2 and Ι3 of ∆ ABC is called
the excentral or excentric triangle.
(i)
∆ ABC is the pedal triangle of the ∆ Ι1 Ι2 Ι3.
(ii)
Its angles are
(iii)
π C
π A π B
& − .
− , −
2 2
2 2 2 2
A
Its sides are 4 R cos ,
2
B
C
4 R cos
& 4 R cos .
2
2
Page : 20 of 29
Circumcentre (O)
98930 58881
(i)
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
12. The Distances of The Special Points from Vertices and Sides of
Triangle:
Ι Ι1 = 4 R sin
(v)
Incentre Ι of ∆ ABC is the
orthocentre of the excentral
∆ Ι1 Ι2 Ι3.
Page : 21 of 29
A
;
2
B
C
Ι Ι2 = 4 R sin ; Ι Ι3 = 4 R sin .
2
2
(iv)
∴
A
A
B
B 
C
C

 2 sin cos   2 sin cos   2 sin cos 
2
2
2
2
2
2





= 8R3 .
A
B
C
cos . cos . cos
2
2
2
A
B
C
A
B
C
= 64R3 sin sin sin
∵
r = 4R sin sin sin
2 22
2
2
2
2
ΙΙ1 . ΙΙ2 . ΙΙ3 = 16R r
Hence Proved
ΙΙ1 + Ι2Ι3 = ΙΙ2 + Ι3Ι1 = ΙΙ3 + Ι1Ι2
2
(ii)
2
2
2
2
2
a2
A
A
+ a2 cosec2
=
A
A
2
2
sin 2 cos 2
2
2
2
2 A
2 A
16 R sin
. cos
A
A
2
2
2
2
2
∵
a = 2 R sinA = 4R sin
cos
∴
ΙΙ1 + Ι2Ι3 =
= 16R
A
A
2
2
2
2
sin
. cos
2
2
2
Similarly
we can prove ΙΙ22 + Ι3Ι12 = ΙΙ32 + Ι1Ι22 = 16R
Hence ΙΙ12 + Ι2Ι32 = ΙΙ22 + Ι3Ι12 = ΙΙ32 + Ι1Ι22
Self Practice Problem :
π
1.
In a ∆ABC, if b = 2 cm, c = 3 cm and ∠A =
, then find distance between its circumcentre and
6
incentre.
∵
Ans.
ΙΙ1 + Ι2Ι3 = a2 sec2
2
2
2 − 3 cm
Exercise -1
Part : (A) Only one correct option
1.
2.
In a triangle ABC, (a + b + c) (b + c − a) = k. b c, if :
(A) k < 0
(B) k > 6
(C) 0 < k < 4
In a ∆ABC, A =
(A) 6 3 cm
(D) k > 4
2π
9 3
, b – c = 3 3 cm and ar (∆ABC) =
cm 2. Then a is
3
2
(B) 9 cm
(C) 18 cm
(D) none of these
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
(i) Distance between circumcentre and orthocentre
OH2 = R2 (1 – 8 cosA cos B cos C)
(ii) Distance between circumcentre and incentre
A
B
C
sin
sin
) = R2 – 2Rr
OΙ2 = R2 (1 – 8 sin
2
2
2
(iii) Distance between circumcentre and centroid
1
OG2 = R2 – (a2 + b2 + c2)
9
Example :
In Ι is the incentre and Ι1, Ι2, Ι3 are the centres of escribed circles of the ∆ABC, prove that
(i)
ΙΙ1. ΙΙ2 . ΙΙ3 = 16R2r
(ii)
ΙΙ12 + Ι2Ι32 = ΙΙ22 + Ι3Ι12 = ΙΙ32 + Ι1Ι22
Solution.
(i)
∵
We know that
A
B
C
ΙΙ1 = a sec , ΙΙ2 = b sec
and ΙΙ3 = c sec
2
2
2
C
A
B
∵
Ι1Ι2 = c. cosec , Ι2 Ι3 = a cosec
and Ι3Ι1 = b cosec
2
2
2
A
B
C
∵
ΙΙ1 . ΙΙ2 . ΙΙ3 = abc sec
sec .sec
........(i)
2
2
2
∵
a = 2R sin A, b = 2R sinB and c = 2R sinC
∴
equation (i) becomes
A
B
C
sec
sec
∵
ΙΙ1. ΙΙ2 . ΙΙ3 = (2R sin A) (2R sin B) (2R sinC) sec
2
2
2
98930 58881
15. Distance Between Special Points :
(A) cos (B – C)
(B) sin (B – C)
b2 − c 2
is equal to
2a R
(C) cos B – cos C
(D) none of these
If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ (= PR), then the angle P is
π
π
2π
π
(A)
(B)
(C)
(D)
6
3
3
2
5.
In a ∆ ABC, the value of
(A)
6.
r
R
acosA + bcosB + ccosC
is equal to:
a+b+c
(B)
R
2r
(C)
In a right angled triangle R is equal to
s+r
s −r
(B)
(A)
2
2
R
r
(C) s – r
(D)
2r
R
(D)
s+r
a
98930 58881
4.
7.
In a ∆ABC, the inradius and three exradii are r, r1, r2 and r3 respectively. In usual notations the value of
r. r 1. r2. r 3 is equal to
abc
(A) 2∆
(B) ∆2
(C)
(D) none of these
4R
8.
In a triangle if r1 > r2 > r3, then
(A) a > b > c
(B) a < b < c
9.
1 1
With usual notation in a ∆ ABC  r + r 
 1 2
where 'K' has the value equal to:
(A) 1
(B) 16
(C) a > b and b < c
1 1
 + 
r

 2 r3 
(D) a < b and b > c
 1 1
KR 3
 +  =
,
r

a2b2 c 2
 3 r1 
(C) 64
(D) 128
10.
The product of the arithmetic mean of the lengths of the sides of a triangle and harmonic mean of the
lengths of the altitudes of the triangle is equal to:
(A) ∆
(B) 2 ∆
(C) 3 ∆
(D) 4 ∆
11.
In a triangle ABC, right angled at B, the inradius is:
AB + BC − AC
AB + AC − BC
AB + BC + AC
(A)
(B)
(C)
(D) None
2
2
2
The distance between the middle point of BC and the foot of the perpendicular from A is :
12.
(A)
13.
− a2 + b2 + c 2
2a
(B)
b2 − c 2
2a
(C)
b2 + c 2
(D) none of these
bc
In a triangle ABC, B = 60° and C = 45°. Let D divides BC internally in the ratio 1 : 3, then,
(A)
2
3
(B)
1
1
(C)
3
(D)
6
sin ∠BAD
=
sin ∠CAD
1
3
14.
Let f, g, h be the lengths of the perpendiculars from the circumcentre of the ∆ ABC on the sides a, b and
a b c
abc
c respectively. If + + = λ
then the value of λ is:
f g h
f gh
(A) 1/4
(B) 1/2
(C) 1
(D) 2
15.
A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length
3, 4 and 5 units. Then area of the triangle is equal to:
(A)
9 3 (1 + 3 )
π
2
(B)
9 3 ( 3 − 1)
π
2
(C)
9 3 (1 + 3 )
2π
2
(D)
9 3 ( 3 − 1)
2π 2
16.
If in a triangle ABC, the line joining the circumcentre and incentre is parallel to BC, then
cos B + cos C is equal to:
(A) 0
(B) 1
(C) 2
(D) none of these
17.
If the incircle of the ∆ ABC touches its sides respectively at L, M and N
and if x, y, z be the circumradii of the triangles MIN, NIL and LIM where
I is the incentre then the product xyz is equal to:
(A) R r2
(B) r R2
(C)
1
R r2
2
(D)
Page : 22 of 29
If R denotes circumradius, then in ∆ABC,
1
r R2
2
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
3.
In any ∆ABC, then minimum value of
(A) 3
(B) 9
r1 r2 r3
r3
(C) 27
(D) None of these
20.
In a acute angled triangle ABC, AP is the altitude. Circle drawn with AP as its diameter cuts the sides
AB and AC at D and E respectively, then length DE is equal to
∆
∆
∆
∆
(A)
(B)
(C)
(D)
3R
2R
4R
R
21.
22.
AA1, BB1 and CC1 are the medians of triangle ABC whose centroid is G. If the concyclic, then points
A, C1, G and B 1 are
(A) 2b2 = a2 + c2
(B) 2c 2 = a2 + b2
(C) 2a2 = b2 + c2
(D) None of these
In a ∆ABC, a, b, A are given and c 1, c2 are two values of the third side c. The sum of the areas of two
triangles with sides a, b, c1 and a, b, c2 is
1
1
(A) b 2 sin 2A (B) a 2 sin 2A
(C) b 2 sin 2A
(D) none of these
2
2
23.
In a triangle ABC, let ∠C =
is equal to
(A) a + b – c
π
. If r is the inradius and R is the circumradius of the triangle, then 2(r + R)
2
[IIT - 2000]
(B) b + c
(C) c + a
(D) a + b + c
24.
Which of the following pieces of data does NOT uniquely determine an acute - angled triangle
ABC (R being the radius of the circumcircle )?
[IIT - 2002]
(A) a , sin A, sin B
(B) a, b, c
(C) a, sin B, R
(D) a, sin A, R
25.
If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side to the perimeter is
(B) 1 : 6
(C) 1 : 2 + 3
(D) 2 : 3
[IIT - 2003]
(A) 3 : (2 + 3 )
26.
The sides of a triangle are in the ratio 1 :
(A) 1 : 3 : 5
27.
3 : 2, then the angle of the triangle are in the ratio
[IIT - 2004]
(C) 3 : 2 : 1
(D) 1 : 2 : 3
In an equilateral triangle, 3 coincs of radii 1 unit each are kept so that they touche each other and also
the sides of the triangle. Area of the triangle is
[IIT - 2005]
(A) 4 + 2 3
28.
(B) 2 : 3 : 4
(B) 6 + 4
3
(C) 12 +
If P is a point on C1 and Q is a point on C2, then
(A) 1/2
(B) 3/4
7 3
4
(D) 3 +
7 3
4
PA 2 + PB 2 + PC 2 + PD 2
equals
QA 2 + QB 2 + QC 2 + QD 2
(C) 5/6
(D) 7/8
29.
A circle C touches a line L and circle C1 externally. If C and C1 are on the same side of the line L, then
locus of the centre of circle C is
(A) an ellipse
(B) a circle
(C) a parabola
(D) a hyperbola
30.
Let " be a line through A and parallel to BD. A point S moves such that its distance from the line BD and
the vertex A are equal. If the locus of S meets AC in A1, and " in A 2 and A3, then area of ∆A1 A2A3 is
(A) 0.5 (unit)2
(B) 0.75 (unit)2
(C) 1 (unit)2
(D) (2/3) (unit)2
Part : (B) May have more than one options correct
31.
In a ∆ABC, following relations hold good. In which case(s) the triangle is a right angled triangle?
(A) r2 + r3 = r1 − r
(B) a2 + b2 + c2 = 8 R2 (C) r1 = s
(D) 2 R = r1 − r
32.
In a triangle ABC, with usual notations the length of the bisector of angle A is :
A
abc cos ec
2 bc cos A
2 bc sin A
2∆ .
A
2
2
2
(A)
(B)
(C)
(D) b + c cos ec 2
b+c
b+c
2R (b + c )
AD, BE and CF are the perpendiculars from the angular points of a ∆ ABC upon the opposite sides,
then :
33.
Page : 23 of 29
is equal to
98930 58881
19.
r
1
A  tan B + tan C 

 is equal to :
If in a ∆ABC, r = , then the value of tan
2
2
2
2 
1
1
(A) 2
(B)
(C) 1
(D) None of these
2
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
18.
(D) Circum radius of ∆DEF =
R
2
34.
The product of the distances of the incentre from the angular points of a ∆ ABC is:
(abc )R
(abc )r
(A) 4 R2 r
(B) 4 Rr2
(C)
(D)
s
s
35.
In a triangle ABC, points D and E are taken on side BC such that BD = DE = EC. If angle
ADE = angle AED = θ, then:
(A) tanθ = 3 tan B
(B) 3 tanθ = tanC
6 tan θ
(C)
= tan A
(D) angle B = angle C
tan2 θ − 9
36.
With usual notation, in a ∆ ABC the value of Π (r1 − r) can be simplified as:
A
(A) abc Π tan
2
1.
(B) 4 r R2
If in a triangle ABC,
(a b c)2
(C)
2
R (a + b + c)
Exercise -2
(D) 4 R r2
cos A + 2 cos C
sin B
=
, prove that the triangle ABC is either isosceles or right
cos A + 2 cos B
sin C
angled.
 A + B
 , prove that triangle is isosceles.
 2 
2.
In a triangle ABC, if a tan A + b tan B = (a + b) tan 
3.

r 
r 
If  1 − 1   1 − 1  = 2 then prove that the triangle is the right triangle.
r2  
r3 

4.
5.
6.
In a ∆ ABC, ∠ C = 60° & ∠ A = 75°. If D is a point on AC such that the area of the ∆ BAD is 3 times
the area of the ∆ BCD, find the ∠ ABD.
The radii r1, r2, r 3 of escribed circles of a triangle ABC are in harmonic progression. If its area is 24 sq.
cm and its perimeter is 24 cm, find the lengths of its sides.
ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that
(
)
2 c 2 −a 2
cos A. cos C =
.
3 ac
7.
8.
9.
Two circles, of radii a and b, cut each other at an angle θ. Prove that the length of the common chord is
2ab sin θ
.
2
a + b 2 + 2ab cos θ
In the triangle ABC, lines OA, OB and OC are drawn so that the angles OAB, OBC and OCA are each
equal to ω, prove that
(i)
cot ω = cot A + cot B + cot C
(ii)
cosec2 ω = cosec2 A + cosec2 B + cosec2 C
In a plane of the given triangle ABC with sides a, b, c the points A′, B′, C′ are taken so that the
∆ A′ BC, ∆ AB′C and ∆ ABC′ are equilateral triangles with their circum radii R a, Rb , Rc; in−radii ra, r b, rc
& ex−radii ra′, rb′ & rc′ respectively. Prove that;
[∑ (3R +6r +2r ′ )] Πtan A
=
3
(i)
Π ra : Π Ra: Π ra ′ = 1: 8: 27
(ii)
r1 r2 r3
a
a
a
10.
2
648 3
The triangle ABC is a right angled triangle, right angle at A. The ratio of the radius of the circle
11.
circumscribed to the radius of the circle escribed to the hypotenuse is, 2 : 3 + 2 . Find the acute
angles B & C. Also find the ratio of the two sides of the triangle other than the hypotenuse.
The triangle ABC is a right angled triangle, right angle at A. The ratio of the radius of the circle
12.
circumscribed to the radius of the circle escribed to the hypotenuse is, 2 : 3 + 2 . Find the acute
angles B & C. Also find the ratio of the two sides of the triangle other than the hypotenuse.
If the circumcentre of the ∆ ABC lies on its incircle then prove that,
13.
(
)
(
)
cosA + cosB + cosC = 2
Three circles, whose radii area a, b and c, touch one another externally and the tangents at their points
of contact meet in a point; prove that the distance of this point from either of their points of contacts
1
 abc  2
 .
is 
a+b+c
Page : 24 of 29
(C) Area of ∆AEF = ∆ cos2A
(B) Area of ∆DEF = 2 ∆ cosA cosB cosC
98930 58881
Perimeter of ∆DEF r
=
Perimeter of ∆ABC R
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
(A)
Page : 25 of 29
2. B
3. B
4. D
5. A
6. B
7. B
8.
A
9.
C
10.
B
11. A
12. B
13. C
14. A
15. A
16. B
17. C
18.
B
19.
C
20.
D
21. C
22. A
23. A
24. D
25. A
26. A
27. B
28.
B
29.
C
30.
C
36.
ACD
31. ABCD 32. ACD
33. ABCD 34. BD 35. ACD
Exercise -2
4. ∠ ABD = 30°
10. B =
π b
5π
,C=
, = 2+ 3
12
12 c
5. 6, 8, 10 cms
11. B =
π b
5π
,C=
, = 2+ 3
12
12 c
TEK
O CLASSES GR
OUP MA
THS BY SUHAA
G SIR PH: (0755)- 32 00 000,
TEKO
GROUP
MATHS
SUHAAG
1. C
98930 58881
Exercise -1