Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Content marketed & distributed by FaaDoOEngineers
...
com
TRIGONOMETRY

BASICS

Some Formulae
1
...





 Cos3A
CosACos  A Cos  A  
4
3

3


3
...






tan+ tan      tan      3 tan 3
3
3



5
...


Sin(A  B)SinA  B  Sin 2 A  Sin 2 B

7
...


CosACos2ACos4A
...
n

9
...

cos
 n
2n  1
2n  1
2n  1
2n  1 2

10
...
Sin  + Sin(  + ) + Sin(  + 2) + ……+ Sin(  + n  1 )

=

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

n
n 1
}Sin
2
2

Sin
2

Content marketed & distributed by FaaDoOEngineers
...
Cos  + Cos (+ ) +Cos ( + 2) + …+ Cos (  + n  1 )

13
...
where   0,  
...
where     , 
 2 2

(f)



2/3

 b2/3



3/ 2

2

 

...

 2

    3 
2 ab
...

 2  2 

The minimum value of a tan + b cot  is

Trigonometrical Equations
(a) SinA = 0  A = n
(b) CosA = 0  A = (2n + 1)/2
(c) TanA = 0

 A = n

(d) Sin A = SinB  A = n + (-1)nB
(e) Tan A = TanB  A = n + B
(f) Cos A = Cos B  A = 2n  B
(g) Sin2A = Sin2B ; Cos2A = Cos2B ; Tan2A = Tan2B

all  A = n  B

Inverse Trigonometry

  
, 
Principal values sin-1x  
 2 2
cos-1x 

[0, π]

sec-1x  [0, π] except π/2

tan-1x  (-π/2, π/2)

  
cosec-1x   ,  except 0
 2 2

cot-1x

If angle is principal one then

sin-1sinx = x

cos-1cosx = x

cosec-1x+sec-1x= π/2 ,

tan-1x + cot-1x= π/2

14
...


sin-1

1
= cosec-1x,
x

cot-1x

 (0, π)


-1 1
, x 0
 tan x
=
cos-1x = sec-1 1/x
-1 1
  tan
, x 0
x


Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

etc

etc

Content marketed & distributed by FaaDoOEngineers
...
sin-1(–x) = – sin-1x

cosec-1 (–x)= –cosec-1x

sec-1 (–x)

tan-1(–x) = –tan-1x

cot-1 (–x) = π – cot-1x

cos-1(–x) = π – cos-1x

17
...
tan-1x - tan-1y = tan-1

= π – sec-1x

= π + tan-1

xy
, x y < -1 , x > 0
1  xy

= - π + tan-1
= π/2 , x y = -1 ,

x> 0

=-π/2

xy
, x y < -1 , x < 0
1  xy

, x y = -1 , x < 0

19
...
sin-1x- sin-1y

if

if

y > 0 &x < 0 & x2 +y2> 1

if

-1  x, y  1 , x + y  0

if

-1  x, ,y  1 , x + y  0

if

-1  x, y  1 , x  y

if

-1  y  0 , 0  x  1 , x  y

if

-1/√2  x  1/√2

21
...
cos-1x _ cos-1y

=cos-1 [ xy  1  x 2
= - cos-1 [ xy  1  x 2

23
...
2cos-1x
25
...
com
= π - sin-1 [3x – 4x3] if 1/2 < x  1

26
...
Projection Formulae
(iv)

a = b cosC + c cosB

(v)

b = c cosA + a cosC

(vi)

c = a cosB + b cosA

28
...
Semi Sum Formulae
(a) Sin

A

2

(s  b)(s  c)
bc

(b) Cos

s(s  a )
A

2
bc

(c) Tan

A

2

(s  b)(s  c)
s(s  a )

30
...
Let R, r, r1, r2, r3 be radii of circumcircle, incircle, excircles of  ABC then
(a) R 

a
b
c


2SinA 2SinB 2SinC

(b) R 

abc
4

(c) r1 


sa

r
r2 


;
sb

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880


s

r3 


sc

Content marketed & distributed by FaaDoOEngineers
...
Regular polygon & radii of the inscribed & circumscribing circles of a regular polygon :
Regular Polygon - It is a polygon whose sides are equal and also its angles are equal
...

(iii)

Area of a regular polygon of n sides =

1 2
 n
2

na cot  R 2 sin
= nr 2 tan
4
n 2
n
n

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

Content marketed & distributed by FaaDoOEngineers
...


BASIC
1
...


(a) 1

(d) N/T

9
...


Smallest
+ve
x
log cos x sin x  log sin x cos x  2
(a) π/2
(c) π /4

satisfying

(b) 2006

If x =

2  2  2  2 cos  then x is

(d) π /6

(a) 2 cos 100

(b) 2 cos 200

(c) 2 cos 400

(b) π /3

sin

(d) N/T

(b) l

(c) 4

then

(c) information insufficient

Number of solutions of sin8x = 1+ tan4x
(a) 0

4
...


[ sinx ] = cosx ,

(b) 2

(c) 3

(b) sin 2 > sin 2

(c) cos 2 > cos 2

Number of solutions of
x  [ 0 , 100 π ] is

(d) 2 cos 800

If tan-1 ( sin2 x ) > 1 then x 
(a) ( 1/√2 , 1)

5
...


(b) ( 0, 1/√2 )

(c) ( -1 , -1/√2 ) U ( 1/√2 , 1) (d) N/T
6
...


If sec  and cosec  are the roots of x2 - px + q
= 0, then

(a) 5

(a) p2 = q(q – 2)

(b) p2 = q(q + 2)

(c) ∞
7
...


cos


3
5
7
9
 cos
 cos
 cos
 cos
11
11
11
11
15

(a) 0

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

(b) – 1/2

Content marketed & distributed by FaaDoOEngineers
...


cos 7  7 cos 5  21cos 3  35 cos 
is equal to
cos 6  6 cos 4  15 cos 2  10
(a) 2 cos 

1
(c) cos 
2
15
...


In a triangle cos 2

A
B
C
A
B
C
cos cos (b) 2 cos sin sin
2
2
2
2
2
2

(c) 2 sin

4x2

If tan A is integral solution of
– 16x + 15 <
0 and cos B is slope of bisector of first
quardrant then sin (A + B) sin (A - B) =
(a) 4/5

26
...


Equation

sin6x

+

(d) N/T
cos6x

=  , has a solution if

1 
(a)    ,1
2 

28
...
terms 
n
n
n
n





Maximum of 3+ sin x    2 cos  x  , is
4
4


(a) 5

(b) – 4/5

(c) 1/5

21
...


A
B
C
sin sin (d) N/T
2
2
2

(c) 1

The equation a sin x + b cos x = c, where | c | >

(d) None of these
...


If sin A + sin B = 0= cos A + cos B , then
cos 2A + cos 2B is equal to

(a) 2 sin

20
...


24
...


(c)

(b) 0

1 
(b) y   ,3
3 

1
 y 3
3

If an angle  is divided into two parts A and B
such that A – B = x and tan A : tan B = k : 1,
then
sin
x
is
k 1
k
sin 
sin 
(a)
(b)
k 1
k 1

2 2

(a) 2

17
...


(d) - 1 + cot 
...


(b) cos 

 
3 
5 
7 

1  cos 1  cos 1  cos 1  cos  is
8 
8 
8 
8


1
(a)
2

16
...


(b) 45

(c) 4 9

(d) N/T

 
Maxmum of 5 cos  +3 cos     + 3 is
3

(a) 5

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

(b) 10

Content marketed & distributed by FaaDoOEngineers
...


(d) –11

(a) 0

(b) 1

(c) 2
31
...


(c) 0

(d) N/T

2

(c) 5
33
...


(d) N/T

41
...


Sum of all roots of cos 6x + sin
x [-, ]

4

x = 1 ,

(d) N/T

1  tan 2 2A tan 2 A

If tan 2  tan  =1, then  is equal to


6

6

(b) n 


6

(d) N/T

General solution of cos x cos 6x = -1
(a) ( 2n + 1) π/2

 tan KA tan A then K is

(a) 1

(b)2

(c) 3

44
...


(c) 2 n 
43
...


(b) {0}

(a) 0

(b) 8

(c) 12

36
...


(d) 3

(b)8

(c)9

(b) 1


(a)  
4

Minimum value of (sin θ + cosec θ )2 +
(sec θ + cos θ )2
(a) 7

No
...


2

(d) 4

(a) 0

Maximum value of 4 Sin2 x + 3 cos2 x +
sin x/2 + cos x/2
(a) 4 +

(b) 2

(c) 3
39
...
cos  + 3
sin2  +2 is
(a) 4 + 10

32
...


4



Maximum of cos2A + cos2B – cos2C

(d) nπ/2

If tan  + tan 4 + tan 7 = tan  tan 4 tan 7
then  is
(a) n/4

x

for

(c) n/12

which
45
...


4



(c) 8

(d) 9

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

Content marketed & distributed by FaaDoOEngineers
...


If the complex numbers sin x + i cos 2x and
cos x – i sin 2x are conjugate to each other,
then x is equal to

54
...


1

(b)  n  
2

(d) N/T

 2 5 
(c) 
, 
 3 6 

x
The number of solutions of 2 cos2   sin 2 x =
2
1

x2 + 2 , 0  x 
is
2
x
(a) 0

48
...


(a) 0

(b) 1

(c) 2
49
...



3

(b)

(c)

General solution for |sin x| = cos x is


3


,0
2

(a)

(d) 32

(a) 2n 

51
...


(b) 24

(c) 28

5
,
6

INVERSE

Number of solution for |sinx| = | cos3x | in
[ - 2π , 2π ] is
(a) 30

50
...


  2 
(b)  ,

2 3 

4
3

(d) N/T

tan-11 + tan-12 + tan-13 = ?
(a) 0
(c) 

58
...


(b) n   / 6
(d) 2 n   / 6

(c) 2 3

(d) 

If the solutions for  of cos p  + cos q  = 0,
p > 0, q > 0 are in A
...
, then the numerically
smallest common difference of A
...
is
(a)
(c)

53
...


(a) 4

60
...


(d) N/T

5
2

The value of cot  cos ec 1  tan 1  is
3
3

(a) 3/17

61
...
com
(c) x > y and y2 = x
62
...


(d) N/T

If sin-1x + sin-1y =  then x100 + y1005 = ?
(a) 0

(b) 2

(c) 1

72
...


If



cos 1x i  0 then

i 1

20

x

i

?

i 1

(a) 0

If

x2 + y2 + z2 = r2 ,
xy
yz
xz
is equal to
tan 1  tan 1  tan 1
zr
xr
yr

74
...


(d) 1
cos-1

(

1  ab
)+
ab

(b) π
(d) N/T

If p > q > 0 & pr < -1 < qr then tan-1

pq
1  pq

qr
rp
+ tan-1
is
1  pr
1  rq

(a) 0
(c) cant say
77
...


(d) N/T

cot -1 (

sin (cot-1(tancos-1x)) = ?
(a) x

69
...


(b) π

(b) 18

(c) 17
68
...


1  ab
) + cot -1 (
ab

1  bc
1  ac
)  cot -1 (
)
bc
ca

(b)  /2

(c) 0

(b) a/b

20

(d) N/T

(a) 

b
b
 1
 1
tan   cos 1  + tan   cos 1 
4 2
a
4 2
a


(c) 2b/a

(b) 1

(c) –1

(d) 

(a) 2a/b


then x2 + y2 + z2 +
2

(a) 0

(b) 1

(c) 2

(d) –1

If sin-1x + sin-1y + sin-1z =

tan 1 x(x  1)  sin 1 x 2  x  1   / 2 , No
...


(b) 0

(c) 3

66
...


(b) 1/2

(b) {-1, 2}

(c) {-1, 1, 0}

64
...


(d) y2 = 1 + x

(b) π
(d) - π
sin-1

If 0 < x < 1 &1 +
x + (sin-1 x )2 + (sin-1 x )3
+ --------∞ = 2 then sin-1x is
(a) π/6

(b) π/3

(c) π /12

(d) π / 4

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

Content marketed & distributed by FaaDoOEngineers
...


Solution

of

the

equation

 1 x
 2x 
3 sin 1 
  4 cos 1 
2
 1 x 2
 1 x 

2






 2x  
2 tan 1 
  for x is
1 x 2  3
(a) x =  3

(b) x = 1/  3

(c) x = 1

 cot r
1

3/ 4




, then the largest angle of a
2
triangle whose sides are 1, sin x, cos x is
If 0 < x <
(a) π/2

86
...




79
...


(d) x

In a triangle ABC, angle A is greater than
angle B
...


Range of

sin-1

x + tan-1 x + cos-1 x is

87
...


88
...


In a triangle ABC, if a,b,c are in A
...
, then
A
B
C
tan
, tan , tan , are in
2
2
2
(a) A
...


(b) G
...


(c) H
...


(c) 27
89
...

A bus on one road is 2 km away from the
intersection and a car on the other is 3 km
away from the intersection ; then the direct
distance between the two vehicles is
(a) 1 km

(d)

7 km

In a Δ ABC, C =

84
...


91
...


If c cos2

(b)1
(d) N/T

A
C 3b
 a cos 2  then a,b,c are in
2
2 2

(a) AP

93
...


(b) a2, b2, c2 –AP

(c) a, b, c –HP

(b) right angled
(d) none of these
...


(b)

(d) N/T

(a)  /4
90
...


(b) 9

(d) N/T

a cot A + b cot B + c cot C = ?

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

Content marketed & distributed by FaaDoOEngineers
...


103
...


If ∆ABC is a right triangle, the value of
 A-BC
 can be equal to
 2ac sin
2


(b) 8R2

1
(c)
2R

1 1
(d) 
r R

98
...
In a triangle ABC with sides a, b, c, r1 > r2 > r3
(which are the ex-radii ), then
(a) a > b > c

(b) a < b < c

(c) a> b and b< c

(d) a< b and b > c
...
If

r r2

r1 r3

then

(d) N/T

If altitudes of a triangle are in AP then sides
are in

(a) A  90

(b) B  90

(c) C  90

(b) G
...


(c) H
...


(d) 2π/3
...
r1 =2 r2 = 3 r3 then

cos( A  C)
If cos 2B 
then tan A,tan B,tan C
cos( A  C)
are in
(a) A
...


(b) π/3

(c) π/2

(d) N/T
...


(a) π/6

(b) 4∆2 / abc

(d) N/T

107
...


(b) HP
(d) AGP

108
...
If b + c = 3a, then the value of cot
(a) 1
(c)

B
C
cot is
2
2

(b) 2

3

(d)

2
...
If cos A + cos B + 2 cos c = 2 then sides are in
(a) AP

(b) GP

(c) HP

(d) N/T

101
...
In  ABC two larger sides are 10 & 9
...
P then 3rd side is
(a) 3 3

(d) N/T

(a) /6

(b) /4

(c) /3

(d) N/T

111
...
In a triangle ABC (Sin A + Sin B + Sin C) (Sin A
+ Sin B - Sin C) = 3 Sin A SinB then angle C

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

Content marketed & distributed by FaaDoOEngineers
...
If two towers of hts b1 and b2 subtend 60 and
30 at mid-point of line joining their feet, then
b1 : b2 =

119
...
A man from the top of a 100 m high tower
sees a car moving towards the tower at an
angle of depression of 30
...
The
distance (in metres) travelled by the car
during this time is
(a) 100

3

100 3
(c)
3

(b)

200 3
3

(d) 200

3

114
...

The breadth of the river is :
(a) 20 m

(b) 30 m

(c) 40 m

(d) 60 m
...
In  ABC, sides a,b,c are in A
...
, a being
smallest, then cosA is
(a)

3c  4b
2c

4c  3b
(c)
2c

(b)

3c  4b
2b

(d) N/T

116
...
Then the measure of the angle C
(a) 30°
(c) 30° or 150°
117
...
Number of solutions of pair of 2 sin2 θ cos2θ = 0 , 2 cos2 θ – 3sinθ = 0 in [ 0, 2π ] is
(a) 1

(b) 0

(c) 2

(d) 4

121
...
If sin2 x – 2 sinx – 1=0 has exactly 6 roots in [
0 , n π ] then minimum of n is
(a) 2

(b) 4

(c) 6

(d) 3

123
...
If sin A – sin B = a & cos A + cos B = b then
(a) a2 + b 2
(c) a2 + b 2 



4 (b) a2 + b 2  4
3

(d) a2 + b 2  2

125
...
For ∆ ABC
sin 2 A + sin
2 cos A cos B cos C is

(a) 0

(b) 3

(a) 0

(d) 5

(c) -2

B + sin

2

C -

(b) 2

(c) 1

2

(d) -1

118
...
Number of points inside or on x2 + y2 =4
satisfying tan4 x + cot 4 x + 1 = 3 sin2 y is
(a) 0

(b) 2

(c) 4

(d) ∞

128
...
com
(a) b sin A  a, A 
(b) b sin A  a, A 
(c) b sin A  a, A 



2

, b  a or b sin a  a, A 



2


...
From the top of a h meter high cliff, angles of
depression of the top and the bottom of a
tower are observed to be 30o and 60o
respectively
...
If the sides of a triangle ABC are in A
...
, and a
is the smallest side, then cos A equal to:

133
...
At a point 15 metres away from the base of a
15 metres high house, the angle of elevation
of the to is
(a) 45 

(b) 30 

(c) 60 

3sin2 A

(a) 15m



(b) 2
(d) 8

134
...
A tree is broken by wind and its upper part
touches the ground at point 10 metres from
the foot of the tree and makes an angle of 45°
with the ground
...
If in ΔABC the line joining the circumcentre
and the incentre is parallel to BC then

(b) 20m



(c) 10 1  2 m

(a) r = R sin A

(b) R = r sin A


3
m
(d) 101 

2 



(c) r = R cos A

(d) R = r cos A

Nishant Gupta, D-122, Prashant vihar, Rohini, Delhi-85
Contact: 9953168795, 9268789880

Content marketed & distributed by FaaDoOEngineers

---