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FEW OF THE GEMS
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1. The sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of
the squares of the lengths of its sides.
2. Given two triangles having one vertex A in common, the other vertices being
situated on two straight lines passing through A then ratio of the areas of these triangles is equal
to the ratio of the products of the two sides of each triangle emanating from the vertex A.
3. The area of the circumscribed polygon is equal to “rp”, where ‘r’ is the radius of the
inscribed circle and ‘p’ its half-perimeter (in particular, this formula holds true for a triangle).
4. The radius of the circle inscribed in a right triangle can be computed by the formula
abc
r
, where a and b are the legs and c is the hypotenuse.
2
5. If a and b are two sides of a triangle,  the angle between them, and ‘I ’ the bisector of this

2abcos
2,
angle, then I 
ab
6. Prove that the distances from the vertex A of the triangle ABC to the points of tangency of
the inscribed circle with the sides AB and AC are equal to p - a (each), where p is the halfperimeter of the triangle ABC, a = | BC |.
7. The legs of a right triangle are a and b. Find the distance from the vertex of the right angle to
the nearest point of the inscribed circle.
8. Given in a triangle ABC are three sides: | BC| = a, | CA | = b, | AB| = c. Find the ratio in
which the point of intersection of the angle bisectors divides the bisector of the angle B.
9. The sum of distances from any point inside an equilateral triangle to its sides is equal to the
altitude of this triangle.
10. Find the area of the quadrilateral bounded by the angle bisectors of a parallelogram with
sides a and b and angle a.
11. Prove that the bisector of the right angle in a right triangle bisects the angle between the
median and the altitude drawn to the hypotenuse.
12. In a triangle ABC, the angle AB C is a. Find the angle AOC, where 0 is the centre of the
inscribed circle.
13. A circle is circumscribed about an equilateral triangle ABC, and an arbitrary point M is taken
on the arc BC. Prove that | AM | = | BM | + | CM |. See the figure. 13.
14. The area of a rhombus is equal to S , the sum of its diagonals is m. Find the side of the
rhombus. See the figure. 14.
15. A square with side a is inscribed in a circle. Find the side of the square inscribed in one of the
segments thus obtained.
16. A circle is circumscribed about a triangle ABC where |BC| = a, ∠B = a, ∠C = β. The bisector
of the angle A meets the circle at a point K. Find |AK|.
17. Find the sum of the squares of the distances from the point M taken on a diameter of a
circle to the end points of any chord parallel to this diameter if the radius of the circle is R,
and the distance from M to the centre of the circle is a.
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PIONEER’S SHORTCUTS
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Illustration: An equilateral triangle has one vertex at the point (0, 0) and another at
(3, 3 ). Find the co-ordinates of the third vertex
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SHORTCUT NO. 2
The straight line ax + by + c = 0 divides the joint of points A (x1 y1 ) and B (x 2 ,y 2 ) in the ratio
on
If ratio is positive then divides internally and if ratio is negative then
divides externally.
SUB-SHORTCUT NO. 1
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If mid points of the sides of a triangle are (x1,y1), (x2, x2) and (x3, x3) then coordinates of the original
triangle are
(x 2  x 3  x 1 , y 2  y 3  y 1 ),(x 3  x 1  x 2 , y 3  y 1  y 2 )
and (x 1  x 2  x 3 , y 1  y 2  y 3 ). (3  x 1  x 2 ,3  y 1  y 2 )
SUB-SHORTCUT NO. 2: If two vertices of a triangle are (x1,y1) and (x2 ,y2) and the co-ordinates of
centroid are (a, p) then co-ordinates of the third vertex are: (3α−x1−x2, 3β−y1−y2)
SUB-SHORTCUT NO. 3: The orthocenter, the nine point centre the centroid and the circum
center therefore all lie on a straight line.
SUB-SHORTCUT NO. 4: If O is orthocenter, N is nine point centre, G is centroid and C is circum
center then to remember it see ONGC (i.e., Oil Natural Gas Corporation) in left of G are 2 and in
right is 1 therefore G divides 0 and C in the ratio 2 : 1 (internally).
SUB-SHORTCUT NO. 5: N is the mid point of O and C
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a1
b1
c1
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1
SUB-SHORTCUT NO. 6 Radius of nine point circle =  Radius of circumcircle
2
Note: 1. The distance between the orthocenter and circumcenter in an equilateral triangle is
zero.
2. The orthocenter of a triangle having vertices (α,β), (β,α) and (α, α) is (α, α).
3. If the orthocenter and centroid of a triangle are respectively (α, β), (γ, δ) then orthocenter will
be (3γ−2α, 3δ−2β).
SUB-SHORTCUT NO. 7 If a1x + b1y + c1= 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 are the sides
of a triangle then the area of the triangle is given by (without solving the vertices)
2
1
a2 b2
c2
2|C1C2C3 |
a3 b3
c3
Where C1 C2, C3 are the cofactors of c1, c2, c3 in the determinant
a2 b2
a b
a b
 (a2b3  a3b2 ) C2  3 3  (a3b1  a1 b3 ) and C3  1 1  (a1b2  a2b1 )
Here, C1 
a3 b3
a2 b2
a1 b1
Illustration:- Find the area of the triangle formed by the straight lines 7x – 2y + 10 =0,
7x + 2y – 10 = 0 and 9x +y + 2 = 0 (without solving the vertices of the triangle).
7x  2y  10  0
Solution: The given lines are: 7x  2y  10  0
9x  y  2  0
7 2
Area of triangle  
where C1 
1
7
2|C1C2C3 |
9
7 2
 7  18  11,
9 1
2
 10
1
2
2
..............(1)
C2 
9 1
 18  7  25
7 2
7
2
and 7
2
1
9
10
 10  10C1  10C2  2C3
2
on
7 2
and C3 
 14  14  28,
7 2
10
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
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10  ( 11)  10  ( 25)  2  28  196
1
196  196
686

 (196)2 

sq units
∴
From (1),
2  11  25  28 275
2| 11  ( 25)  28|
Complex number as a rotating arrow in Argand plane:
Let z = r (cos θ + i sin θ) = reiθ ... (1)
be a complex number representing a point P in the Argand plane.
Then OP = | z| = r
and ∠POX = θ
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Now consider complex number zr = zeiϕ
or
z1=reiθ eiϕ =r.ei(θ+ϕ)
{from (1)}
Clearly the complex number Z 1 represents a point Q in the Argand plane, when
OQ = r and ∠QOX = θ + ϕ

Clearly multiplication of z with ei ϕ rotates the vector OP through angle (j> in anticlockwise sense.

Similarly multiplication of z with e−i ϕ will rotate the vector OP in clockwise sense.
Note: If z1, z2 and z3 are the affixes of the three points A, B and C such that AC = AB and
∠CAB = θ Therefore
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

AB  z2  z1 ,AC  z3  z1


Then AC will be obtained by rotating
through
an angle θ in
AB
 i
AC

ABe
anticlockwise sense and therefore
Or (z3−z1)= (z2−z1)eiθ
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Solution: Here AB = (2  3)2  (0  1)2  2
10
and slope of AB 
= 1 = tan 45° 3-2
32
∠BAX=45°
em
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 z z 
or  3 1   ei
 z 2  z1 
or
(z3 –z1) = (z2-Z1)e'e
ILLUSTRATION: The line joining the-points A (2, 0) and B (3,1) is rotated about A in the
anticlockwise direction through an angle of 15°. Find the equation of the line in the new position.
If B goes to C in the new position, what will be the co-ordinates of C?
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Now line AB is rotated through an angle of 15°
⟹
∠CAX=45° + 15° = 60° and AB = AC = 2
Equation of line AC in parametric form is
x  2  rcos 60 
 Since AC = r = 2
y  0  r sin 60 
1 4 2
3
6
2 in (1), then x  2  2. 
and y  2.

2
2
2
2
x2
1
 cot 600 
Equation of the line AC is
or x 3  y  2 3  0
y
3
 4 2 6 
,
and co-ordinates of C are 
.
 2
2 

Alternative Method:
∵ A = (2,0), B = (3,1), let C = (x,y)
Put r =
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5
i
zC  z A
 e 12
zB  zA
0
zc =2 = (1 + i(cos15 + i sin 15°) or
 3 1
3 1 
 zC  2  (1  i)
i

 2 2
2 2  

 3 1
3 1   3 1
3 1 
 2

 i



 2 2
2 2   2 2
2 2 

1   3

 2

  i
2   2 

4 2  6 

 i
 2 
2


 4 2 6 
∴ C
,

 2
2 

and equation of AC, y − 0 = tan 60° (x −2), x 3  y  2 3  0
ILLUSTRATION: The centre of a square is at the origin and one vertex is A (2,1). Find the
co-ordinates of oilier vertices of the square.
Solution:
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∴ ZA = 2, ZB = 3 + i, zc = x + iy =
∵ A = (2,1) ∴ zA = 2+i Now in triangle AOB, OA=OB,  AOB = 90° =
i

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 ZB  ZA e 2  iz A  2i  1
 B  ( 1,2)
∵ O is the mid point of AC and BD

C  ( 2, 1) and D  (1, 2)

2
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 6x  4y  3  0
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1

y 0 2
x 

3 1
2
  
2 2
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ILLUSTRATION: The extremities of the diagonal of a square are (1,1), (-2, -1). Obtain
the other two vertices and the equation of the other diagonal.
Solution:
Let ABCD be a square
∵ A = (1,1)
∴
Z A=1+i
and let A (-1,-2) and C (3,2) be the given points.
and
C = (−2,−1)
z c = −2 −i
Let B (x,y) be the unknown vertices
∴ AB = BC
1
 1 
then centre of E    ,0   zE  
= >AB2 = BC2
2
 2 
(x +1)2 + (y -2)2 = (x-3)2 + (y-2)2


zB  zE
x = 1…………………..(i)
e 2 i
Now in △AEB, (EA = EB)
z A  zE
In right angled triangle, ABC,
we have (AB2+ BC2 = (AC)2
3 3
3 3

 zB    i then D   1  ,  
2x2 + 2y2 – 4x -8y + 18
2 2
2 2

= (3+1)2 + (2-2)2
 3 3
1 3
y 2−4y =0
B   , ,
D   , 
 2 2
2 2
y =0 , y =4
Hence equation of other diagonal BD is
Hence required vertices of square are (1,0) and
3
(1,4)
0
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Let Q be the reflection of P (4,1) about the
line y = x, then Q = (1, 4)
∵ Q move 2 units along the +ve direction of
x-axis, if new point is R then R = (3, 4).
If
R(3,4) = R (z1)
when z1 = (3 + 4i)
then R'(x, y) = R'(z2)


z2=z1eiπ/4
 ROR '  
4

i 
 1




 (3  4i) cos  isin   (3  4i)

4
4
2

 2
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Co-ordinates of Q is (1, 4).
Given that Q move 2 units along the positive
direction of x-axis.
∴ Co-ordinates of R is (x1 + 2, y1) or R(3, 4)
If OR makes an angle θ, then
4
4
3
tan  
 sin   and cos  
3
3
5

After rotation of let new position of R is R' and
4
Pioneer Smart Solution
(Use of complex number)
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Solution: Let Q (x1 , y1 ) be the reflection of P
about the line y = x. Then
x1  1 
 ................(1)
y 1  1
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ILLUSTRATION: The point (4,1) undergoes the following three transformations successively :
(i) Refection about the line y = x .
(ii) Translation through distance 2 units along the positive direction of x-axis.
(iii) Rotation through an angle π/ 4abnui the origin in the anticlockwise direction. Then find the
co-ordinates of the final position.
 1 7 
,
Hence new co-ordinates are  

 2 2 .

1
 1
 
cos  
sin   , 
 OR' 
2
 2
 
R'

1
 1

cos  
sin   
 OR'sin 
2
 2


  3
4   3
4 
R ' 5


 ,5

  5 2 5 2   5 2 5 2 
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i,e.,
7i 
 1



2
 2
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4 OR=OR' = 32  42  5
∴ OR' makes an angle (π/ 4 + θ)with x-axis.
Co-ordinates of R'





 OR'cos     ,OR'sin     
4

4




 1 7 
R ' 
,

2 2

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“OBJECTIVE QUESTIONS”
1. (1  i)  (1  i) 
(a)28
(b)0
6
(1  i)6        ..............................(ii)
adding (i)and (ii), the terms those exist at even
places get cancelled
 (1  i)6  (1  i)6  2[6 C0 6 C2i2 6 C4 i4 6 C6 i6 ]
 2[6 C0 6 C2 6 C4 6 C6 ] using i2  1
2. The square root of −5−12i is
(a) (3  2i)
(b) (2  3i)
 5  2

36

 5  2 (9)(4)
5  2 (3i)2 (2)2
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 9  4  2.2(3i)  (  3i)2  22  2.2( 3i)
5  12  (2  3i)2

5  12i  (2  3i)
(c) (2  3i)
(d) (3  2i)
“Pioneer Smart Solution”.
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Sol. (c)
5  12i  5  2(6i)
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6
 2(0)
C2 6 C4
0
Alternative method II:



1  i  2  cos  isin 
4
4


1  i  2 cis
4
 
1  i  2 cis   
 4
 3
3 
 3  

( 2)6 cis  cis     ( 2)6 2cos   0
2
2

 2  ,

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(1  i)        .............................(i)
z  1  cos   isin 
2
2
z2/3  cos  isin
3
3
2k


2



 2k  2 
z2/3  cos 
  isin 

3
3




2

4

z2/3  cis ,cis ,cis 2
3
3
2
2 

 cis ,cis  2   ,cis 2
3
3 

2
 2 
 cis , cis    ,cis 0
3
 3 
2 2
Arg z2/3 
,  ,0
3
3
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6
“Pioneer Smart Solution”:
em
a
Sol. (D) Alternative Method I:
Binomial coefficients in (1+x)6 are
6
C0 ,6 C1 ,6 C2 ,6 C36C4 ,6 C5 ,6 C6
(d) 1
m
(c)−1
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Every complex number possesses its two
square roots. From these two both, one or none
may or may not be in the choice, so be careful
about it.
Fact
 |z |  Rez
|z |  Rez 
a  ib   
i
 a is b  0
2
2


 13  ( 5)
13  ( 5) 
5  12i 
i
  (2  3i)
2
2


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6
(c) Im (z) = 0
Sol. (c)
6
“Pioneer Smart Solution”:
6
6
 3 i  3i
z 

 2   2 

 

Fact: Using fact every complex number a +ib
for which |a:b|= 1: 3 or 3 :1 can be
expressed in terms of i, , 2 .
6
6
6
(a)0
6
 12  2  2  0i
i
8
8
 isin
then Re(    2   3   4   5 ) equals
11
11
(b) −1/2
(c) 1/2
ath
4. If   cos

em
a
 3  i   3  i   1  i 3   1  i 3 
 


  

 2   2  
2
2

 
 
 


    ( 2 )6
 i
 Im(z)  0
Sol. (b)
zz
2
(d) None of these
“Pioneer Smart Solution”:
ee
rm
Def: Rez=
6
s.c
o
6
 3 1  3 i 

z
 2   2 

 

i
 i
 2cos 
e e
z  2  2  0i
tic
6
(d) Re z > 0, Im (z)< 0
m
6
 3 i  3i
3. If z  
then

 2   2 

 

(a)Re z =0
(b) Re z, Im z >0
8
8
Given 2  cos
 isin
3
4
5
               11    11   11
 R e (   2  3   4   5 ) 
2    cos 8   isin 8  1
2
3
4
5
ww
w.
pi
on
10
1  11
1
1
1
1
1 
n
2
3
4
5
so


 0 (sum of 11th roots of units)
           2  3  4  5 

1
2
    
n 0
1
[ 6  7   8   9  10   4  3   2    1]
2 2
z  z sum of 11th root of units  1
1
now
Re(z)



1
10
5
2
2
2
[1    .........     ]
22
 1
1 1  1  11
1
 5 
[0  5 ]  
2 
2
2   1
2
 2
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(c) 64
Sol. (c)
z 1i 3
(d) None of these
“Pioneer Smart Solution (i)”
Z 1 i 3
 1  i 3 
z  1i 3
2
 
  2  2

2
3
4
5 2
3 6
6 C0 6 C1 i 3 6 C1 i 3 6C3 i 3 6 C3 i 3 6 C5 6 i 3 6C
i
6
z  ( 2)6 ( 2 )6  26 12
6


 
 
 
 
 
 
6
6
 (6 C0 6 C2(3) 6 C4 9 6 C0 27)  i 3(6 C1  36 C3  96 C5 ) ∴ z  2  64
s.c
o
6
m
5. If z  1  i 3 then z6 equals
(a) 32
(b) −32
“Pioneer Smart Solution (ii)”
 (1  15 3  135  27)  i 3(60  60)
tic
z 1i 3
1
3
z  2  i

2
2 




z  2 cos  isin 
3
3

em
a
 (136  72)  i 3(0)  64
z  2ei/3
 z6  26 e2i  26
(a) x – axis
(b) y – axis
(c)on a circle
“Pioneer Smart Solution”:
z  7i
1
Given
z  7i
⟹ z lies on the right bisector of the line
segment connection the points 7i, −7i.
∴ z lies on x-axis
Hence z lies on real axis.
ww
w.
pi
on
ee
rm
Sol. (a)
z  7i
1
Given
z  7i
|z  7i|| z  7i|
| x  i(y  7)|| x  i|(y  7)|
 x 2  (x  7)2  x 2  (y  7)2 (after taking
absolute value squaring both side)
 (y  7)2  (y  7)2  0
 28y  0
(Equation of x-axis)
y 0
⟹ z lies on x – axis.
z  7i
1
z  7i
(d)the line y = 7
ath
6. The complex number z  x  iy which satisfying the equation
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH
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“Pioneer Smart Solution”:
Using Fact: If vertices of a ∆ABC are (a,1/a),
(b,1/b),(c,1/c) then coordinate of
 1

, abc 
orthocenter is  
 abc

s.c
o
Sol. (c): Orthocenter is point of intersection of
altitudes drawn from one vertex to opposite
side. In order to determine the co-ordinate of
orthocenter we need the equations of altitudes
m
7. If the vertices of a triangle is (4,1/4)(3,1/3),(1,1) then orthocenter of the triangle is
1 
 1


 1

(a)  ,12 
(b)  12, , 
(c)   , 12 
(d) None of these
 12

 12 
 12

tic
∴coordinate of orthocenter

  1
1


, 4(1)(3)    , 12 

 (4)(3)(2)
  12
em
a
1
1
 1 / 3
Slope of BC 3
31
ee
rm
ath
∴ Equation of AD is y− 1/4 = 3(x − 4)
12x − 4y = 47
Equation of BF is
12x − y = 11
By solving (i) and (ii) we get the co-ordinate of
orthocenter
 1

(x,y)    ,12 
∴
 12

8. The angle between the pair of tangents drawn from the point (2,4) to the circle x2  y2  4 is
3
4
(a) tan 1  
(b) tan 1  
(c)900
(d) None of these
8
3
ww
w.
pi
on
Sol. (b): Equation of pair of tangent to a circle is
T2 = SS1 where T = xx1+yy1−4
S1=x 21+ y 21−4
S = x2 +y2 −4
 (2x  4y  4)2  (x 2  y 2  4)(4  16  4)
4(x  2y  2)2  16(x 2  y 2  4)
x 2  4y 2  4xy  8y  4x  4  4x 2  4y 2  16
3x 2  4xy  8y  20  0 
 PT  4
radius = OT =2
2 1
now tan   
4 2
1
2tan 
2 4
and tan   tan2 

2
1  tan  1  1/ 4 3
2.
ax 2  2hxy  by 2  2gx  2fy  c
2 h2  ab 2  2

ab
3
1
  tan (4 / 3)
tan  
“Pioneer Smart Solution”:
The length of tangent
 PT  x12  h21  a2  22  42  4
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193
3
(b)
193
6
(c)
129
3
(d) None of these
on
ee
rm
ath
em
a
tic
s.c
o
(a)
m
9. The radius of the circle 3x 2  3y 2  9x  8y  4  0 is
“SPECIAL CONSTANTS”
1.
π = 3.14159 26535 89793 23846 2643 .....
2.
1

e = 2.71828 18284 59045 23536 0287 ….. = lim  1   natural base of logarithms.
n 
n

ww
w.
pi
n
3.
4.
5.
6.
7.
8.
9.
2 = 1.41421
3 = 1.73205
5 = 2.23606
3
2 = 1.25992
3
3 = 1.44224
5
2 = 1.14869
5
3 = 1.24573
35623 73095 0488 ……
08075 68877 2935 ……
79774 99789 6964 ……
1050 ……
9570 ……
8355 ……
0940 ……
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH
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ee = 15.15426 22414 79241 90 ……
log 102 = 0.30102 99956 63981 19521 37389 ……
log 103 = 0.47712 12547 19662 43729 50279 ……
log 10 e = 0.43429 44819 03251 82765 ……
log 10π = 0.49714 98726 94133 85435 12683 ……
loge 10 = ln 10 = 2.30258 50929 94045 68401 7991 ……
loge 2 = ln 2 = 0.69314 71805 59945 30941 7232 ……
loge 3 = ln 3 = 1.09861 22866 68109 69139 5245 ……
γ = 0.57721 56649 01532 86060 6512 …… = Euler’s constant =
21.
tic
1 1
1


lim  1    ......   ln n 
x 
2 3
n


m
eπ = 23.14069 26327 79269 006 ……
π e = 22.45915 77183 61045 47342 715 ……
s.c
o
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
eγ = 1.78107 24179 90197 9852 ……
e = 1.64872 12707 00128 1468 ……
23.
1
π = 1   = 1.77245 38509 05516 02729 8167 ….. Where Γ is the gamma function.
1 radian = 180o /π = 57.29577 95130 8232 ……….o
1o = π /180 radians = 0.01745 32925 19943 29576 92 ……… radians.
GREEK ALPHABET
Aα
B β
ath
26.
27.
1
Γ   = 2. 67893 85347 07748 ……
3
1
Γ   = 3.62560 99082 21908 ……
4
alpha
beta
gamma
ν nu
Ξ ξ xi
N
Oo
omicron
π
Pρ
σ
T τ
υ
pi
rho
ww
w.
pi
Γγ
Δ δ delta
E  ε epsilon
ee
rm
25.
2
on
24.
em
a
22.
Zζ
Hη
zeta
eta
Θ θ , theta
Iι
iota
K κ kappa
Λ λ lambda
M μ mu
sigma
tau
upsilon
Φ  , φ phi
X χ chi
Ψ ψ psi
Ω ω omega
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH
(all x)
(all x)
tic
x
)
2
s.c
o
(all x)
em
a
(|x| <
(|x| < 1)
(|x| < 1)
ww
w.
pi
on
ee
rm
ath
“Special Power Series”
x2 x3
xr
ex = 1 +x +
+…

 .... 
2! 3!
r!
x3 x5 x 7
( 1)r x 2r 1
sin x = x +…
   ... 
3! 5! 7!
(2r  1)!
x2 x 4 x6
( 1)r x 2r
cos x = 1 +…

  ... 
2! 4! 6!
(2r )!
x3 2x5 17x7


tan x = x +
+…
3
15
315
1 x3 1.3 x5 1.3.5 x7
-1



sin x = x +
2 3 2.4 5 2.4.6 7
1.3.5 ... (2n  1) x 2n  1
+…
... 
2.4.6 .... (2n ) 2n  1
2n 1
x3 x5 x7
n x
-1


 ...  (1)
 ...
tan x = x 3
5
7
2n  1
n
x 2 x3 x 4
n 1 x
 ...  (  1)
 n(1 + x) = x –  
+…
2 3
4
n
x3 x5 x 7
x2n 1
Sinh x = x +
+…

  ... 
3! 5! 7!
(2n  1)!
x2 x4 x6
x 2n
coshx = 1 +
+…
   ... 
2! 4! 6!
(2n)!
x3 2x5 17x7


tanh x = x +…
3! 15
315
1 x3 1.3 x5 1.3.5 x7
–1



sinh x = x 2 3 2.4 5
2.4.6 7
1.3.5....(2n  1) x2n1
… + (-1)n
+…
2.4.6 ... 2n 2n  1
x3 x5 x 7
x 2n 1
–1
tanh x = x +
+…


 ...
3
5
7
2n  1
π
1 1 1 . 3 1 1 .3 .5 1
1

. 
.  ........
2
2 3 2 . 4 5 2. 4 . 6 7
π2
1
1
1
 2  2  2  .......
6
1
2
3
m
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(– 1 < x  1)
(all x)
(all x)
(|x| <
x
)
2
(|x| < 1)
(|x| < 1)
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH
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e-x
cosh x =
+ )/2
tanh x = sinh x/cosh x
sechx = 1 / cosh x
coth x = cosh x/ sinh x = 1/tanh x
cosh ix = cos x
cos i x = cosh x
cosh2 A – sinh2 A = 1
sech2 A = 1 – tanh2 A
cosech2 A = coth2 A- 1
sinh (x) = (ex – e-x )/2
ath
sinh ix = i sin x
sin ix = i sinh x
em
a
cosech x = 1/sinh x
tic
“Hyperbolic Identities”
(ex
s.c
o
m
π2 1
1
1
1
 2  2  2  2  .............
12 1
2
3
4
2
π 1
1
1
 2  2  2  ..........
8 1
3
5
π
1 1 1
 1     ..........
4
3 5 7
1
1
1
1
= 


..........
2 1 .3 3.5 5.7
“PHYSICAL AND ASTRONOMICAL CONSTANTS”
σ
C1
C2
ww
w.
pi
εo
μo
ee
rm

K
G
Speed of light in vacuum
Elementary charge
Neutron rest mass
Proton rest mass
Electron rest mass
Planck’s constant
Dirac’s constant (= h/2 π )
Boltzmann’s constant
Gravitational constant
Stefan-Boltzmann constant
First Radiation Constant (= 2 π hc2)
Second Radiation Constant (= hc/k)
Permittivity of free space
Permeability of free space
Avogadro constant
Gas constant
on
c
E
mn
mp
me
h
NA
R
2.998 × 108 m s–1
1.602 × 10-19 C
1.675  10-27 kg
1.673  10-27 kg
9.110  10-31 J s
6.626  10-34 Js
1.055  1034 J s
1.381  10-23 J K-1
6.673  10-11 N m2 kg-2
5.670  10-8 J m-2 K-4 s-1
3.742  10-16 J m2 s-2
1.439  10-2 m K
8.854  10-12 C2 N-1 m-2
4 π  10-7 H m-1
6.022  1023 mol-1
8.314 J K-1 mol-1
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH
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First check your Concept
=
±√
?

Why sum of

why sin 30 = ?

Why cos 45 =

Why centroid of triangle whose vertices are ( ,
of an A.P. is [2 + ( − 1) ]?
?
), ( ,
?

Why distance between ( ,

Why sum of interior angles of triangle is 1800?

Why volume of sphere is

Why curved surface area of cylinder is 2

Why area of circle is r 2 ?

Why cos is negative in second quadrant?

Why sum of roots of quadratic equation is b / a ?
) and ( ,
) is (
) is
−
) −(
−
) ?
ee
rm
ath
?
), ( ,
tic
√
em
a
,
m
+ = 0 are
Why roots of
s.c
o
+

ℎ?
………….100’s of many more why’s.,
ww
w.
pi
on
come to PIONEER EDUCATION
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