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AJAY PARMAR
GROUP TUITION
Chapter-9 : Trigonometry
C
Side opposite to angle 
Trigonometric ratios for angle  in a right angled triangle:
If in ABC, mB = 90 and mA =  then

A Side adjacent to angle  B


Side opposite to angle θ
=
Hypotenuse
Side adjacent t o angle θ
cos  =
=
Hypotenuse
sin  =
BC
AC
AB
AC


Side opposite to angle θ
BC
=
Side adjacent t o angle θ
AB
Hypotenuse
AC
=
Side opposite to angle θ
BC
Hypotenuse
AC
sec  =
=
Side adjacent t o angle θ
AB
cosec  =
AB
Side adjacent t o angle θ
=
Side opposite to angle θ
BC

tan  =


The trigonometric ratios remain same for a given angle and even if the lengths of the sides of given
triangle are multiplied to some common number, the value of these ratios do not change.
The short names of sine, cosine, tangent, cosecant, secant and cotangent are sin, cos, tan, cosec, sec and
cot respectively.
A is generally represented as A.
(sin )2, (cos )2, (tan )2, (cosec )2, (sec )2 and (cot )2 are written as sin2, cos2, tan2, cosec2,
sec2 and cot2 respectively.
Inter – relationships of trigonometric functions:

tan  =


Identities related to inter – relationships of trigonometric ratios:
tan   cot  = 1, cosec   sin  = 1, sec   cos  = 1.
Values of trigonometric ratios for 0, 30, 45, 60 and 90.



sin θ
1
=
;
cosθ
cot θ
cosec  =
sin

1
;
sin θ
0
0
cos
1
tan
0
cosec
Not
defined
sec
1
cot
Not
defined
sec  =
30
1
2
3
2
1
3
2
45
1
2
1
2
1
2
cot  =
1
cos θ
60
3
2
1
2
and
cot  =
cos θ
1
=
tan θ
sin θ
90
1
0
3
Not
defined
2
1
3
2
Not
defined
2
3
3
1
1
0
3
Trigonometric ratios for complementary angles:
For every   R, and 0 <  < 90
sin (90 – ) = cos 
cos (90 – ) = sin 
tan (90 – ) = cot 
cot (90 – ) = tan 
cosec (90 – ) = sec 
sec (90 – ) = cosec 
c.g.road
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AJAY PARMAR
GROUP TUITION
Identities:
For every   R, and 0 <  < 90
sin2 + cos2 = 1 ……(i)
sin2 = 1 – cos2 and cos2 = 1 – sin2
Dividing (i) by cos2 we get
and
Dividing (i) by sin2 we get
sin 2 θ
2
cos θ

cos 2 θ
=
2
cos θ
sin 2 θ
1
2
2
cos θ
sin θ

cos 2 θ
2
sin θ
=
1
sin 2 θ

tan  + 1 = sec ,
1 + cot  = cosec2,
2
2
sec  – tan  = 1 and
cosec2 – cot2 = 1 and
sec2 – 1 = tan2
cosec2 – 1 = cot2
Thus we can express each trigonometric ratio in terms of other trigonometric ratios as follows:
sin  = 1  cos 2 θ
and
cos  = 1  sin 2 θ

cosec  = 1  cot 2 θ
and
cot  = cos ec 2θ  1

sec  = 1  tan 2 θ
and
tan  = sec 2 θ  1

cosec  =
and
sec  =

tan  =
and
cot  =
2
2
1
1  cos θ
sin θ
2
1  sin 2 θ
2
1
1  sin 2 θ
cos θ
1  cos 2 θ
Exercise – 9.1
1. In ABC, mA = 90. If AB = 5, AC = 12 and BC = 13, find sin C, cos C, tan B, cos B and sin B.
sin C =
cos C =
tan B =
AB
BC
AC
BC
AC
AB
5
13
12
=
13
12
=
5
B
=
13
C
12
5
cos B =
sin B =
AB
BC
AC
BC
5
13
12
=
13
=
A
2. In ABC, mB = 90. If BC = 3, AC = 5, find all the six trigonometric ratios of A.
ABC, mB = 90. BC = 3, AC = 5
C
By Pythagoras theorem
AB2 + BC2 = AC2
5
3
2
2
2
AB + 3 = 5
AB2 + 9 = 25
A
4
B
AB2 = 25 – 9 = 16
AB = 4
sin A =
cos A =
tan A =
BC
AC
AB
AC
BC
AB
3. If cos A =
4
5
=
=
=
AC
BC
AC
sec A =
=
AB
AB
cot A =
=
BC
3
5
4
5
3
4
cosec A =
, find sin A and tan A.
AB
4
= AB = 4k and AC = 5k
5
AC
By Pythagoras theorem
(4k)2 + BC2 = (5k)2
16k2 + BC2 = 25k2
c.g.road
5
3
5
4
4
3
C
Let ABC, mB = 90.
cos A =
=
A
5
4
3
B
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BC2 = 25k2 – 16k2 = 9k2
BC = 3k
sin A =
tan A =
BC
AC
BC
AB
= 3k = 3
=
5k
3k
4k
=
5
3
4
4. If cosec  = 13 , find tan  and cos .
5
Let ABC, mB = 90 and mA = 
cosec  =
AC
13
=
AC = 13k and BC = 5k
5
BC
By Pythagoras theorem
AB2 + (5k)2 = (13k)2
 AB2 + 25k2 = 169k2
AB2 = 169k2 – 25k2 = 144k2
AB = 12k
12
BC
12k
=
=
AB
5k
5
12
AB
12k
cos  =
=
=
AC
13k
13
1
If cos B = , find other five
3
C
13
5

A
12
B
tan  =
5.
trigonometric ratios.
B
Let ABC, mA = 90.
cos B =
AB
BC
=
1
AB = k and BC = 3k
3
By Pythagoras theorem
AB2 + AC2 = BC2
(k)2 + AC2 = (3k)2
k2 + AC2 = 9k2
AC2 = 9k2 – k2 = 8k2
AC = 2 2 k
sin B =
AC
2 2
2 2k
=
=
3k
3
BC
sec B =
BC
AB
=
3k
=3
k
3
C
1
A
2 2
tan B =
cot B =
AC
AB
AB
AC
=
=
2 2k
=2 2
k
k
2 2k
=
cosec B =
BC
AC
3k
=
2 2k
=
3
2 2
1
2 2
6. In ABC, mA = 90 and if AB : BC = 1 : 2 find sin B, cos C and tan C.
Let ABC, mA = 90. AB : BC = 1 : 2 
B
AB = k and BC = 2k
2
By Pythagoras theorem
1
AB2 + AC2 = BC2
(k)2 + AC2 = (2k)2
C
A
3
k2 + AC2 = 4k2
AC2 = 4k2 – k2 = 3k2
AC = 3 k
AC
AC
3
3
3k
3k
=
=
cos C =
=
=
2k
2k
2
2
BC
BC
5sinθ  2cosθ
4
7. If tan  = , find the value of
.
3
3sinθ  cosθ
5 sin θ  2 cos θ
3sin θ  cos θ
sin B =
c.g.road
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tan C =
AB
AC
=
k
3k
=
1
3
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AJAY PARMAR
GROUP TUITION
 sin θ   cos θ 
5
  2

cos θ   cos θ 
= 
[Dividing both numerator and denominator by cos   0]
 sin θ   cos θ 
3


 cos θ   cos θ 
4
20  6
5   2
26
3
5 tan θ  2
20  6

=
=
= 3 =
=
12

3
4
3 tan θ  1
12  3
9
 
3   1
3
3
2sinθ  3cosθ
13
8. If sec  = , find the value of
.
5
5cosθ  4sinθ
Let ABC, mB = 90 and mA = 
sec  =
AC
AB
=
13
5
A
AC = 13k and AB = 5k
13
By Pythagoras theorem
AB2 + BC2 = (13k)2
(5k)2 + BC2 = 169k2
25k2 + BC2 = 169k2
BC2 = 169k2 – 25k2 = 144k2
BC = 12k
C

5
B
12
12
5
AB
5k
and
cos  =
=
=
AC
13k
13
13
 12   5 
48  15
2   3 
39
13   13 
2 sin θ  3 cos θ
48  15


=
= 13 =
=–
25  48
5 cos θ  4 sin θ
25  48
23
 5   12 
5   4 
13
 13   13 
1
9. If sin B = , prove that 3cos B – 4 cos3 B = 0.
2
C
sin  =
BC
AC
=
12k
13k
=
Let ABC, mA = 90.
OR
AC
1
sin B =
=
2
BC
2
AC = k and BC = 2k
B
3
1
2
 B = 30
1
sin B =
A
cos B = cos 30 =
3
2
By Pythagoras theorem
AB2 + AC2 = BC2
AB2 + (k)2 = (2k)2
 AB2 + k2 = 4k2
AB2 = 4k2 – k2 = 3k2
AB = 3 k
AB
BC
 cos B =
=
3k
3
=
2
2k
Now
3
 3




 – 4  3  = 3 3 – 4  3 3  = 3 3 – 3 3 = 0 = RHS

 8 
 2 
2
2
2
 2 




LHS = 3cos B – 4 cos3 B = 3 
10. If tan A = 3 , verify that (1) sin2 A + cos2 A = 1, sec2 A – tan2 A = 1 and 1 + cot2 A = cosec2 A.
A
Let ABC, mB = 90.
OR
tan A =
BC
AB
= 3
2
tan A =
BC = 3 k and AB = k
sin A = sin 60 =
C
By Pythagoras theorem
c.g.road
3  A = 60
1
3
B
1
3
; cos A = cos 60 = ;
2
2
2
cosec A = cosec 60 =
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; sec A = sec 60 = 2
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AJAY PARMAR
GROUP TUITION
AC2 = AB2 + BC2
and cot A = cot 60 =
1
3
AC = (k) + ( 3 k)
AC2 = (k)2 + ( 3 k)2
AC2 = k2 + 3k2
AC2 = 4k2
AC = 2k
2
2
2
sin A =
BC
AC
=
sec A =
AC
AB
=
3k
3
=
;
2
2k
2k
= 2;
k
cos A =
AB
AC
=
k
2k
cot A =
AB
BC
=
k
=
3 1

4 4
1
;
2
=
3k
=
cosec A =
1
3
AC
BC
=
2k
3k
=
2
3
.
Now
2
2
2
(1) LHS = sin A + cos A =
 3   1 2

 
 2   2 


(2) LHS = sec2 A – tan2 A =(2)2 –
 3
2
=
3 1
4
=
4
= 1 = RHS
4
= 4 – 3 = 1 = RHS
2
2
 1 
 2 
1
4
3 1
 1 +
=
= =   = cosec2A = RHS
3
3
3
 3
 3
(3) LHS =1 + cot2 A = 1 + 
11. If cos  =
2 2
verify that tan2  – sin2  = tan2  . sin2 .
3
Let ABC, mB = 90 and mA = 
cos  =
AB
AC
=
2 2
AB = 2 2 k and AC = 3k
3
By Pythagoras theorem
AB2 + BC2 = AC2
(2 2 k)2 + BC2 = (3k)2
8k2 + BC2 = 9k2
BC2 = 9k2 – 8k2 = k2
BC = k
tan  =
BC
AB
=
k
2 2k
=
C
3
1

A
1
sin  =
and
2 2
2
2 2
BC
AC
=
B
k
1
=
3
3k
2
 1  1
1 1
1
9 8
    =  =
LHS = tan  – sin = 
=
…(i)
72
8 9
72
 2 2  3
2
2
2
2
 1  1
1 1
1
   =  =
RHS = tan  sin  = 
…(ii)
8 9
72
 2 2  3
2
2
By (i) and (ii) LHS = RHS
12. In ABC, mB = 90. AC + BC = 25 and AB = 5, determine the values of sin A, cos A and tan A.
ABC, mB = 90. AC + BC = 25
A
Let BC = x
AC = 25 – x
25 – x
AB2 + BC2 = AC2
5
2
2
2
5 + x = (25 – x)
25 + x2 = 625 – 50x + x2
C
B
x
50x = 600
 x = 12 BC = 12
AC = 25 – 12 = 13
Thus AB = 5, BC = 12 and AC = 13
c.g.road
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AJAY PARMAR
sin A =
BC
AC
GROUP TUITION
= 12 ;
cos A =
13
AB
5
= ;
AC
13
tan A = BC = 12
AB
5
13. In ABC, mC = 90 and mA = mB. (1) Is cos A = cos B? (2) Is tan A = tan B? (3) Will the
other trigonometric ratios of A and B be equal? B
In ABC, mC = 90 and mA = mB.
BC = AC = k suppose
k
2k
By Pythagoras theorem
AC2 = AB2 + BC2 = x2 + x2 = 2x2
k
A
C
AC = 2 x
(1) cos A =
AB
=
AC
(2) tan A =
BC
AB
=
x
2x
x
x
1
=
2
=1
and
cos B =
BC
AC
=
and
tan B =
AB
BC
=
x
=
2x
1
2
cos A = cos B.
x
= 1 tan A = tan B.
x
(3) The other trigonometric ratios will also be same as the length of adjacent sides, opposite sides and
hypotenuse of A and B are also same.
14. If 3cot A = 4, examine whether
Let ABC, mB = 90
3cot A = 4
cot A =
4
3
1  tan 2 A
= cos2 A – sin2 A.
2
1  tan A
C
5
AB
4

=
3
BC
3
 AB = 4k and BC = 3k
A
4
By Pythagoras theorem
AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 16k2 + 9k2 = 25k2
AC = 5k tan A =
BC
AB
=
3k
4k
=
B
3k
4k
3
AB
4
BC
3
, sin A =
=
= and cos A =
=
=
5
AC
5
AC
5k
5k
4
2
3
9
16  9
1  
1
1  tan 2 A
7
4

16
LHS =
=
=
= 16 =
…(i)
2
2
9
16

9
25
1  tan A
3
1

1  
16
16
4
2
2
4
3
16 9
16  9
7
RHS = cos2 A – sin2 A =      =
=
=
…(ii)

5
5
25
25
25
25
 by (i) and (ii) LHS = RHS
15. If pcot  = q, examine whether
pcot  = q cot  =
p2  q2
psinθ  qcosθ
= 2 2.
psinθ  qcosθ
p q
q
p
q
p2  q2
 sin θ   cos θ 
p  q 
p
  q

p2  q2
p  q cot θ
p sin θ  q cos θ
sin θ   sin θ 
p
 p =
LHS =
= 
=
=
= 2 2 = RHS
2
2
p sin θ  q cos θ
p  q cot θ
 sin θ   cos θ 
q
p q
p q
p
  q

p  q 
p
p
 sin θ   sin θ 
 
16. State whether the following are true or false. Justify your answer.
1. sin  =
sin  =
3
for some angle having
2
Side opposite to angle θ
Hypotenuse
measure .
False.
In a right angled triangle the hypotenuse is greater than the side opposite to angle .
c.g.road
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AJAY PARMAR
GROUP TUITION
Side opposite to angle θ
<1
Hypotenuse
3
3
sin  < 1 but > 1  sin  = is not possible.
2
2
2
2. cos  = , for some angle having measure .
3
Side adjacent t o angle θ
cos  =
Hypotenuse

True.
In a right angled triangle the side adjacent to angle  is less than hypotenuse.
Side opposite to angle θ
<1
Hypotenuse
2
2
cos  < 1 and < 1  cos  = is possible.
3
3
5
3. cosec A = , for some measure of angle A.
2
Hypotenuse
cosec A =
Side opposite to angle A

True.
In a right angled triangle the hypotenuse is greater than side opposite to angle A.

Hypotenuse
5
5
> 1 and > 1  cosec A = is possible
Side opposite to angle A
2
2
4. The value of tan A is always less than 1.
False.
Side opposite to angle A
tan A =
Side adjacent t o angle A
But in a right angled triangle the side opposite to A can be equal, greater than or smaller than the
side adjacent to A.
tan A can be 1, greater than 1 or smaller than 1.
The given statement is false.
5. sec B =
sec B =
3
5
for some B.
False.
Hypotenuse
Side adjacent t o angle B
But in a right angled triangle the hypotenuse is greater than side adjacent to angle B.

3
3
Hypotenuse
> 1 but < 1  sec B = is not possible.
Side adjacent t o angle B
5
5
6. cos  = 100 for some angle having measure of . False.
cos  =
Side adjacent t o angle θ
Hypotenuse
In a right angled triangle the side adjacent to angle  is less than hypotenuse.

Side opposite to angle θ
<1
Hypotenuse
cos  < 1 but 100 > 1  cos  = 100 is not possible.
c.g.road
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AJAY PARMAR
GROUP TUITION
Exercise – 9.2
1. Verify:
1. cos 60 = 1 – 2 sin2 30 = 2 cos2 30 – 1 = cos2 30 – sin2 30.
cos 60 =
1
2
…(i)
2
1
1
1
1
1 – 2 sin2 30 = 1 – 2   = 1 – 2   = 1 – =
2
2
2 cos 30 – 1


= 2  3 
 2 
2
4
2
2
…(ii)
3 2
3
3
1
– 1 = 2   – 1 = – 1 =
= …(iii)
4
2
 3
1
cos2 30 – sin2 30 =   –  
 2 
2
2
2
2
2
3 1
2
1
3 1
= = …(iv)
 =
2
4
4 4
4
=
by (i), (ii), (iii) and (iv)
cos 60 = 1 – 2 sin2 30 = 2 cos2 30 – 1 = cos2 30 – sin2 30.
2. sin 60 = 2 sin 30 cos 30
LHS = sin 60 =
3
2
…(i)
1
2
RHS = 2sin 30 cos 30 = 2 

3
3
=
2
2
…(ii)
2
2
By (i) and (ii) LHS = RHS
3. sin 60 =
2tan30
1  tan 2 30
3
LHS = sin 60 =
2
2 tan 30
RHS =
=
1  tan 2 30
…(i)
 1 

2
 3
 1 

1  
 3
2
2
= 3 = 3 =
3 1
1
1
3
3
2 3
3
3
 =
=
4
2
3 4
3
…(ii)
By (i) and (ii) LHS = RHS
4. cos 60 =
1  tan 2 30
1  tan 2 30
1
LHS = cos 60 =
2
RHS =
1  tan 2 30
1  tan 2 30
=
…(i)
 1 

1  
 3
2
 1 

1  
 3
2
3 1
1
3 = 3 = 2 = 1
=
3 1
1
2
4
1
3
3
1
…(ii)
By (i) and (ii) LHS = RHS
5. cos 90 = 4 cos3 30 – 3 cos 30
LHS = cos 90 = 0
…(i)
3
RHS = 4 cos 30 – 3 cos 30 =
 3
4  
 2 
2


 3
 = 4  3 3   3 3 = 3 3 – 3 3 = 0 …(ii)



2
2
2
 2 
 8 
– 3 
By (i) and (ii) LHS = RHS
2. Evaluate:
1.
sin30  tan45  cosec60
sec30  cos60  cot45
sin 30  tan 45  cos ec60
sec 30  cos 60  cot 45
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AJAY PARMAR
GROUP TUITION
1
2
1
2
3
=
2 1
 1
3 2
3 2 3 4
=
2 3
4 3 2 3
=
3 34
3 34
=
3 3 4 3 3 4

3 3 4 3 3 4
=
(3 3 ) 2  2(3 3) (4)  (4) 2
(3 3 )  (4)
2
2
=
27  8 3  16
43  8 3
=
11
27  16
2 3
5cos 60  4sec 2 30  tan 2 45
2
2.
sin 2 30  cos 2 30
2
 2 
1
5 16
67
15  64  12
  (1) 2
5   4
 1
2
2
2
5 cos 60  4 sec 30  tan 45
67
3
2

4
3
12
=
=
=
= 12 =
2
2
2
1

3
4
1
3
2
12
sin 30  cos 30

 1   3 
  
4
4
4
4
 2   2 
2
3. 2sin2 30 cot 30 – 3 cos2 60 sec2 30
2
1
1
= 2  ( 3 )  3 
2
2
2
2
 2 
3
1
 1  4 

 = 2 ( 3 )  3   =
1 =
2
4
 4  3 
 3
3 2
2
4. 3cos2 30 + sec2 30 + 2 cos 0 + 3sin 90 – tan2 60
3cos2 30 + sec2 30 + 2 cos 0 + 3sin 90 – tan2 60
2
 3  2 
 
  2(1)  3(1)  ( 3 ) 2
  3
2


 
2
= 3
9 4
3
4
27  16  24
67
= 3     + 2 + 3 – 3 =  + 2 =
=
4 3
4
12
3
12
3. In ABC, mB = 90, find the measure of the parts of the triangle other than the one which are
A
given.
1. mC = 45, AB = 5
? 5
?
mB = 90, mC = 45
 mA = 180 – (90 + 45) = 45 mA = 45
45
?
C
B
Now as mA = mC  AB = BC and also AB = 5 BC = 5
Also sin C = sin 45 =

1
2
=
AB
AC
5
AC
AC = 5 2
2. mA = 30, AC = 10
mB = 90, mA = 30
 mC = 180 – (90 + 30) = 60 mC = 60
Now sin C = sin 60 =

3
2
=
AB
AC
AB
10
and sin A = sin 30 =
and
AB = 5 2
1
BC
=
10
2
C
10
BC
AC
?
?
30
A
?
B
BC = 5
3. AC = 6 2 , BC = 3 6
mB = 90, AC = 6 2 , BC = 3 6
 cos C =
BC
AC
=
3 6
6 2
=
6 2
3
mC = 30
2
 mA = 180 – (90 + 30) = 60
c.g.road
A
?
?
C
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B
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X MATHS gb
10
AJAY PARMAR
cos A =
AB
AC

GROUP TUITION
1
AB
=
 AB = 3 2
2
6 2
A
4. AB = 4, BC = 4.
mB = 90, AB = BC = 4  mA = mC = 45
Also AC2 = AB2 + BC2 = 42 + 42 = 16 + 16 = 32 = 16  2
AC = 4 2
? 4
?
?
4
C
B
4. In a rectangle ABCD, AB = 20, mBAC = 60, calculate the length of side BC and diagonals AC
and BD .
mB = 90 in ABC and mA = 60
D
C
AB
AC
20
60 =
AC
cos A =
cos

1
20
=
AC = 40
AC
2
?
?
But in rectangle ABCD AC = BD BD = 40
BC
AC
BC
3

2
40
?
Also sin A =
sin 60 =
=
60
BC
BC = 20 3
40
A
B
20
5. If  is a measure of an acute angle and cos  = sin , find the value of 2tan  + sin2  + 1.
cos  = sin   = 45 (  is an acute angle)
2tan2  + sin2  + 1
2
2
 1 
1
1
7
6 1
= 2tan 45 + sin 45 + 1 = 2(1) +   + 1 = 2(1) + + 1 = 3 + =
=
2
2
2
2
 2
2
2
2
2
 1  tan 2α   2tanα  2
 
6. If  is a measure of an acute angle and 3 sin  = 2 cos , prove that 
 =1
2 
2
 1  tan α 
2
2
4
sin α
= tan  =
tan2  =
9
cos α
3
3
5
4
94
=
=
and
1 + tan2 
9
9
9
 1  tan α 
3 sin  = 2 cos  
 1 – tan2  = 1 –
2
 1  tan 2 α   2 tan α  2
 
LHS = 
 =
2 
2
 1  tan α 
 4 
 
5  3 
  
 13   3 
 
 9 
4
9
=
94
9
=
13
9
2
2
2
=
 1  tan α 
2
 5   2 2  
 
 9   3
 3   3 
  

 9   9 
=1+
2
=
 5  4 9 
    
 13   3 13 
2
2
2
5
12
25  144 169
25 144
=      =
=
=
= 1 = RHS

169 169
169
169
 13   13 
7. If A = 30 and B = 60, verify that
(1) sin(A + B) = sin A cos B + cos A sin B and
(2) cos (A + B) = cos A cos B – sin A sin B.
A = 30 and B = 60  A + B = 30 + 60 = 90
(1) LHS = sin(A + B) = sin 90 = 1 …(i)
RHS = sin A cos B + cos A sin B
= sin 30 cos 60 + cos 30 sin 60
=
1
2

1
2
+
4
3
3
1 3
1 3

= + =
= = 1 …(ii)
4 4
4
2
2
4
by (i) and (ii) LHS = RHS
(2) cos (A + B) = cos A cos B – sin A sin B.
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AJAY PARMAR
LHS
RHS
GROUP TUITION
= cos (A + B) = cos 90 = 0 …(i)
= cos A cos B – sin A sin B
= cos 30 cos 60 – sin 30 sin 60
3
3
1
1
 – 
2
2
2
2
3
3
=
–
= 0 …(ii)
4
4
=
 by (i) and (ii) LHS = RHS
8. If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values
of sin 15 and cos 15.
sin (A – B) = sin A cos B – cos A sin B
cos (A – B) = cos A cos B + sin A sin B
Let A = 45 and B = 30
 sin (45 – 30) = sin 45 cos 30 – cos 45 sin 30
cos (45 – 30) = cos 45 cos 30 + sin 45 sin 30
 sin 15 =
 sin 15 =
 sin 15 =
 sin 15 =
 sin 15 =
1
2
3
1
1
–

2
2
2

3
2 2
–
cos 15 =
1
cos 15 =
2 2
3 1
cos 15 =
2 2
3 1

2 2
2
cos 15 =
2
6 2
4
1
2
3
1
1
+

2
2
2

3
2 2
+
1
2 2
3 1
2 2
3 1

2 2
2
2
6 2
4
cos 15 =
9. State whether the following are true or false. Justify your answer.
1. The value of sin  increases as  increases from 0 to 90.
True.
1
2
sin 0 = 0, sin 30 =
, sin 45 =
1
2
, sin 60 =
3
and sin 90 = 1
2
Thus as the value of  increases from 0 to 90, the value of sin  also increases.
2. sin  = cos  for all values of .
sin  = cos  is possible only for 45 for an acute angle.
Hence sin   cos  for all values of .
False.
3. cos (A + B) = cos A + cos B.
Substituting the value of A and B as 45
False.
LHS = sin 90 = 0 and RHS = cos 45 + cos 45 =
1
2
+
1
=
2
2
2
Thus LHS  RHS.
4. tan A is not defined for A = 90.
tan 90 =
sin 90
cos 90
=
True.
1
which is not defined.
0
5. The value of cot  increases as  increases from 0 to 90.
cot 0 = not defined, cot 30 = 3 , cot 45 = 1, cot 60 =
1
3
False.
and cot 90 = 0
Thus as the value of  incrases from 0 to 90, the value of cot  decreases.
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AJAY PARMAR
12
GROUP TUITION
Exercise – 9.3
1. Evaluate:
1.
cos 18
sin 72
=
cos 18
cos 18
=
=1
cos(90  72)
cos 18
2. tan 48 – cot 42 = tan 48 – tan (90 – 42) = tan 48 – tan 48 = 0
3. cosec 32 – sec 58 = sec(90 – 32) – sec 58 = sec 58 – sec 58 = 0
4.
cos 70
+ cos 59 . cosec 31
sin 20
cos 70
=
+ sin (90 – 59) . cosec 31
sin 20
cos 70
=
+ sin 31 . cosec 31
cos(90  20)
cos 70
=
+ 1 [ sin  . cosec  = 1]
cos 70
=1+1=2
5. sec 70 sin 20 – cos 20 cosec 70
sec 70 sin 20 – cos 20 cosec 70
= sec 70 cos(90 – 20) – cos 20 sec (90 – 70)
= sec 70 cos 70 – cos 20 sec 20
=1–1
[ cos  . sec  = 1]
=0
6. cos(40 – ) – sin (50 + ) +
cos 2 40  cos 2 50
sin 2 40  sin 2 50
cos 2 40  sin 2 (90  50)
= cos[90 – (40 – )] – sin (50 + ) + 2
sin 40  cos 2 (90  50)
= cos[90 – 40 + ] – sin (50 + ) +
= cos[50 + ] – sin (50 + ) + 1
=0+1=1
7.
cos 2 40  sin 2 40
sin 2 40  cos 2 40
cos 70
cos 55 cos ec35

sin 20 tan 5 tan 25 tan 45 tan 65 tan 85
sin(90  70)
cos 55  sec(90  35)

=
sin 20
tan 5  tan 25  1  cot(90  65)  cot(90  85)
sin 20
cos 55  sec55
=

sin 20 tan 5  tan 25  cot 25  cot 5
cos 55  sec55
=1+
(tan 5  cot 5)(tan 25  cot 25)
= 1 + 1 [ cos  . sec  = 1, tan  . cot  = 1]
=2
8. cot 12 . cot 38 . cot 52 . cot 60 . cot 78
= cot 12 . cot 38 . tan (90 – 52) .
= cot 12 . cot 38 . tan 38 .
1
3
.
1
=11
c.g.road
3
=
tan (90 – 78)
3
tan 12
= (cot 12 . tan 12) (cot 38 . tan 38) .
1
.
1
[But tan  . cot  = 1]
3
1
3
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AJAY PARMAR
9.
sin18
+ 3 (tan 10 tan 30 tan 40 tan 50 tan 80)
cos 72
1
sin 18
=
+ 3 [tan 10 
 tan 40  cot (90 –
sin(90  72)
3
=
sin 18
+
sin 18
3[
1
3
50) cot (90 – 80)]
 (tan 10  cot 10)  (tan 40  cot 40)]
= 1 + (1)  (1)
=1+1=2
10.
GROUP TUITION
[ tan  . cot  = 1]
cos 58 sin 22
cos 38 cos ec52


sin 32 cos 68 tan 18 tan 35 tan 60 tan 72 tan 55
sin(90  58)
sin 22
cos 38  sec(90  52)


=
sin 32
sin(90  68) tan 18  tan 35  3  cot(90  72)  cot(90  55)
=
sin 32 sin 22
cos 38  sec38
[But cos  . sec  = 1, tan  . cot  = 1]


sin 32 sin 22
3  (tan 18  cot 18)(tan 35  cot 35)
=1+1–
1
3 1
=2–
1
3
=
2 3 1
3

3
3
=
6 3
3
2. Prove the following:
1. sin 48 sec 42 + cos 48 cosec 42 = 2
LHS = sin 48 sec 42 + cos 48 cosec 42
= cos (90 – 48) sec 42 + cos 48 sec (90 – 42)
= cos 42 sec 42 + cos 48 sec 48
=1+1
[cos  . sec  = 1]
= 2 = RHS
2.
sin70 cosec20
– 2cos 70 cosec 20 = 0

cos20
sec70
sin 70 cos ec 20
LHS =
– 2cos 70 cosec 20

cos 20
sec 70
cos(90  70) sec(90  20)
=
– 2cos 70 sec (90 –

cos 20
sec 70
cos 20 sec70
=
– 2 cos 70 sec 70

cos 20 sec 70
20)
= 1 + 1 – 2(1) [cos  . sec  = 1]
= 2 – 2 = 0 = RHS
3.
tan(90  A)cotA
2
– cos2 A = 0
cosec A
tan(90  A) cot A
LHS =
– cos2 A
cos ec 2 A
cot A  cot A
=
– cos2 A
2
cos ec A
cot 2 A
2
=
– cos A
cos ec 2 A
cos 2 A
=
– cos2 A
2
2
sin A cos ec A
= cos2A – cos2A
= 0 = RHS
4.
[sin  . cosec  = 1]
cos(90  A)sin(90  A)
= sin2 A
tan(90  A)
cos(90  A) sin(90  A)
LHS =
tan(90  A)
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X MATHS gb
AJAY PARMAR
14
GROUP TUITION
= sin A  cosA
cot A
sin A  cosA
=
cos A
sin A
sin A
=
1
sin A
= sin2 A = RHS
3. Express the following in terms of trigonometric ratios of angles having measure between 0 and 45:
1. sin 85 + cosec 85 = cos (90 – 85) + sec (90 – 85) = cos 5 + sec 5
2. cos 89 + cosec 87 = sin (90 – 89) + sec (90 – 87) = sin 1 + sec 3
3. sec 81 + cosec 54 = cosec (90 – 81) + sec (90 – 54) = cosec 9 + sec 36
4. For ABC, prove that
A C
B
 = cot
2
 2 
1. tan 
In ABC, A + B + C = 180
A + C = 180 – B
B
AC
= 90 –
2
2
AC
B

 tan 
 = tan  90  
2

 2 
B
AC
 tan 
 = cot
2
2


BC
A
cos 
 = sin
2
 2 

2.
In ABC, A + B + C = 180
B + C = 180 – B
A
BC
= 90 –
2
2
B

C
A


 cos 
 = cos  90  
2
 2 

A
BC
 cos 
 = sin
2
2



5. If A + B = 90 prove that
tanA tanB  tanA cotB
= sec A.
sinA sec B
A + B = 90
 B = 90 – A
LHS =
tanA tanB  tanA cotB
sinA sec B
=
tanA  cot(90  A)  tanA  tan (90  A)
sinA  cosec(90  A)
=
tanA  cotA  tanA  tan A
sinA  cosecA
[But tan  . cot  = 1, sin  . cosec  = 1]
=
1  tan 2 A
1
[But 1 + tan2A = sec2A]
= sec 2 A = sec A = RHS
c.g.road
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GROUP TUITION
6. If 3 is the measure of an acute angle and sin 3 = cos ( – 26) then find the value of .
sin 3 = cos ( – 26)
 sin 3 = sin [90 – ( – 26)]
 sin 3 = sin (90 –  + 26)
 sin 3 = sin (116 – )
3 = 116 –  4 = 116  = 29
7. If 0 <  < 90, and sin  = cos 30, then obtain the value of 2tan2 – 1.
sin  = cos 30
sin  = sin (90 – 30)
 sin  = sin 60
 = 60
2tan2 – 1 = 2( 3 )2 – 1 = 2(3) – 1 = 6 – 1 = 5
8. If tan A = cot B, prove that A + B = 90, where A and B are measure of acute angles.
tan A = cot B
 tan A = tan (90 – B)
A = 90 – B
A + B = 90
9. If sec 2A = cosec(A – 42), where 2A is the measure of an acute angle, find the value of A.
sec 2A = cosec(A – 42)
sec 2A = sec[90 – (A – 42)]
sec 2A = sec(90 – A + 42)
 sec 2A = sec(132 – A)
2A = 132 – A 3A = 132 A = 44
10. If 0 <  < 90 and sec  = cosec 60, find the value of 2cos2  – 1.
sec  = cosec 60
sec  = sec (90 – 60)
 sec  = sec 30
 = 30
2cos  – 1 =
2
c.g.road
 3
2(  
 2 
2
3
3 2
3
1
)2 – 1 = 2   – 1 = – 1 =
=
4
2
2
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GROUP TUITION
Exercise – 9
Prove the following using trigonometric identities: (1 – 19)
1. cos2  +
1
=1
1  cot 2θ
1
2
LHS = cos  +
1  cot 2 θ
= cos2  + sin2  [
= cos2  +
1
cos ecθ
1
cos ec 2θ
= sin ]
= 1 = RHS
2. 2sin2 + 3sec2 + 5cot2 + 2cos2 – 3tan2 – 5cosec2 = 0
LHS = 2sin2 + 3sec2 + 5cot2 + 2cos2 – 3tan2 – 5cosec2
= 2sin2 + 2cos2 + 3sec2 – 3tan2 – 5cosec2 + 5cot2
=2(sin2 + cos2) + 3(sec2 – tan2) – 5(cosec2 – cot2)
=2(1) + 3(1) – 5(1)
= 2 + 3 – 5 = 0 = RHS
3.
4.
1
1
= 2 cosec2 

1  cosθ 1  cosθ
1
1
2
2
1
1
LHS =
=
=
= 2 = 2cosec2


1  cos θ 1  cos θ
1  cos θ 1  cos θ
1  cos 2 θ
sin θ
secθ  1
tanθ  sinθ
=
secθ  1
tanθ  sinθ
 1

sin θ
sin θ
 1
 sinθ
tan θ  sinθ
 cos θ  = sec θ  1 = RHS
LHS =
= cos θ
=
sin θ
tan θ  sin θ
 1

sec θ  1
 sin θ
sin θ
 1
cos θ
 cos θ 
1  sinθ
= sec  – tan 
1  sinθ
5.
LHS =
1  sin θ
1  sin θ
(1  sin θ)2
(1  sin θ)2
1  sinθ 1  sinθ
=
=

2
1  sin θ 1  sin θ
1  sin θ
cos 2θ
secθ  tanθ
cosecθ  cotθ
=
cosecθ  cotθ
secθ  tanθ
sec θ  tan θ
LHS =
cos ecθ  cot θ
sec θ  tan θ
secθ  tan θ cosecθ  cot θ
=


cosecθ  cot θ cosecθ  cot θ secθ  tan θ
=
6.
= RHS
=
1 sin θ
1
sin θ
=
= sec + tan = RHS

cos θ cos θ
cos θ
cosecθ  cot θ

cosec 2θ  cot 2 θ secθ  tan θ
= 1  cos ecθ  cot θ
secθ  tan θ
cos ecθ  cot θ
=
= RHS
secθ  tan θ
cotθ  cosecθ  1
7.
= cosec  + cot 
cotθ  cosecθ  1
cot θ  cos ecθ  1
LHS =
cot θ  cos ecθ  1
cos ecθ  cot θ  (cos ec 2θ  cot 2 θ)
=
[cosec2  – cot2  = 1]
cot θ  cos ecθ  1
=
sec 2θ  tan 2 θ
c.g.road
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GROUP TUITION
cos ecθ  cot θ  (cos ecθ  cot θ)(cos ecθ  cot θ)
cot θ  cos ecθ  1
(cos ecθ  cot θ)[1  (cos ecθ  cot θ)]
=
1  cos ecθ  cot θ
(cos ecθ  cot θ)(1  cos ecθ  cot θ)
=
1  cos ecθ  cot θ
=
= cosec  + cot  = RHS
8. (sin  + cosec )2 + (cos  + sec )2 = 7 + tan2  + cot2 
LHS = (sin  + cosec )2 + (cos  + sec )2
= sin2  + 2 sin cosec  + cosec2  + cos2  + 2 cos sec + sec2 
= sin2  + cos2  + 2 + 2 + sec2  + cosec2 
[sin cosec  = 1 and cos sec = 1]
2
2
= 1 + 2 + 2 + 1 + tan  + 1 + cot 
[sec2  = 1 + tan2 and cosec2  = 1 + cot2]
2
2
= 7 + tan  + cot  = RHS
9. 2sec2  – sec4  – 2 cosec2  + cosec4  = cot4  – tan4 
LHS = 2sec2  – sec4  – 2 cosec2  + cosec4 
= sec2  (2 – sec2 ) – cosec2  (2 – cosec2 )
= (1 + tan2 )[2 – (1 + tan2 )] – (1 + cot2 )[2 – (1 + cot2 )]
= (1 + tan2 )(2 – 1 – tan2 ) – (1 + cot2 )(2 – 1 – cot2 )
= (1 + tan2 )(1 – tan2 ) – (1 + cot2 )(1 – cot2 )
= (1 – tan4 ) – (1 – cot4 )
= 1 – tan4  – 1 + cot4 
= cot4  – tan4  = RHS
10. (sin  – sec )2 + (cos  – cosec )2 = (1 – sec  cosec )2
LHS = (sin  – sec )2 + (cos  – cosec )2
= (sin  – sec )2 + (cos  – cosec )2
= sin2  – 2 sin sec  + sec2  + cos2  – 2 cos cosec + cosec2 
= (sin2  + cos2 ) – 2(sin sec  – cos cosec) + (sec2  + cosec2 )
sin θ cos θ   1
1 
= (1) – 2 


 +
2
2 
 cos θ
sin θ 
 cos θ
sin θ 
2 
 sin θ  cos θ 

 +  sin θ  cos θ 
= 1 – 2 

 cos 2 θ sin 2 θ 
 cos θ sin θ 


2
2
2
1
1




 + 
2
2
cos
θ
sin
θ


 cos θ sin θ 
= 1 – 2 
= 1 – 2 seccosec + sec2  cosec2 
= (1 – sec  cosec )2 = RHS
11.
2
sinA  cosA sinA  cosA
=

2
sinA  cosA sinA  cosA
sin A  cos 2 A
sin A  cos A sin A  cos A
LHS =

sin A  cos A sin A  cos A
(sin A  cos A) 2  (sin A  cos A) 2
=
(sin A  cos A)(sin A  cos A)
=
=
=
=
2
1  2cos 2 A
sin 2 A  cos 2 A  2 sin A cos A  sin 2 A  cos 2 A  2 sin A cos A
(sin 2 A  cos 2 A)
2
11
=
= RHS(1)
2
2
2
sin A  cos A
sin A  cos 2 A
2
2
1  cos 2 A  cos 2 A
c.g.road
=
1  2 cos 2 A
= RHS(2)
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GROUP TUITION
12. tanθ  cotθ = sec2  – cosec2  = tan2  – cot2 
sinθ.cosθ
tan θ  cot θ
LHS =
sin θ cos θ
sin θ cos θ

= cos θ sin θ
sin θ cos θ
sin 2 θ  cos 2 θ
= cos θ sin θ
sin θ cos θ
=
=
sin 2 θ  cos 2 θ
sin 2 θ cos 2 θ
sin 2 θ

cos 2 θ
sin 2 θ cos 2 θ sin 2 θ cos 2 θ
1
1
=

cos 2 θ sin 2 θ
= sec2  – cosec2 
= (1 + tan2 ) – (1 + cot2 )
= 1 + tan2  – 1 – cot2 
= tan2  – cot2 
13.
secθ  tanθ
= 1 – 2sec  tan 
secθ  tanθ
sec θ  tan θ
LHS =
sec θ  tan θ
sec θ  tan θ sec θ  tan θ
=

sec θ  tan θ sec θ  tan θ
=
+ 2 tan2 
(sec θ  tan θ) 2
sec 2 θ  tan 2 θ
= sec2 – 2sec  tan  + tan2 
= 1 + tan2  – 2sec  tan  + tan2 
= 1 – 2sec  tan  + 2tan2  = RHS
14. sec 2θ  cosec 2θ = tan  + cot 
LHS = sec 2 θ  cosec 2θ
= tan 2 θ  1  1  cot 2 θ
= tan 2 θ  2  cot 2 θ
= tan 2 θ  2  cot 2 θ { tancot = 1)
= tan 2 θ  2  cot 2 θ
= tan + cot = RHS
15.
1
1
1
1
=


cosecA  cotA sinA
sinA cosecA  cotA
1
1

LHS =
cos ecA  cot A sin A
cosec 2 A  cot 2 A
– cosec A
cosecA  cot A
(cosecA  cot A)(cosecA  cot A)
=
– cosec A
cosecA  cot A
RHS =
1
1

sin A cosecA  cot A
cosec 2 A  cot 2 A
cosecA  cot A
(cosecA  cot A)(cosecA  cot A)
= cosec A –
cosecA  cot A
=
= cosec A –
= cosec A + cot A – cosec A
= cot A ……(i)
= cosec A – cosec A + cot A
= cot A
……(ii)
c.g.road
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GROUP TUITION
by (i) and (ii)
1
1
1
1
=


sin A cos ecA  cot A
cos ecA  cot A sin A
tanθ
cotθ
16.
= 1 + tan  + cot  = 1 + sec  cosec 

1  cotθ 1  tanθ
tan θ
cot θ
LHS =

1  cot θ 1  tan θ
1
tan θ
=
 tan θ
1
tan θ  1
1
tan θ
tan θ
1
=

tan θ  1 tan θ(tan θ  1)
tan θ
=
tan 2 θ
1

tan θ  1 tan θ(tan θ  1)
=
tan 3 θ  1
tan θ(tan θ  1)
=
(tan θ  1)(tan 2 θ  tan θ  1)
tan θ(tan θ  1)
=
tan 2 θ  tan θ  1
tan θ
=
tan 2 θ tan θ
1


tan θ tan θ tan θ
= tan  + 1 + cot 
= 1 + tan  + cot  = RHS(1)
sin θ cos θ 
= 1 + 


 cos θ sin θ 
 sin 2 θ  cos 2 θ 

= 1 + 

 cos θ sin θ 
1

= 1 + 

cos
θ
sin
θ


= 1 + seccosec = RHS(2)
17. sin4  – cos4  = sin2  – cos2  = 2sin2  – 1 = 1 – 2 cos2 
LHS = sin4  – cos4 
= (sin2  – cos2 )(sin2  + cos2 )
= (sin2  – cos2 )(1)
= sin2  – cos2  = RHS(1)
= sin2  – (1 – sin2 )
and
= (1 – cos2) – cos2
= sin2  – 1 + sin2 
and
= 1 – cos2 – cos2
= 2sin2  – 1 = RHS(2)
and
= 1 – 2 cos2  = RHS(3)
18. tan2 A – tan2 B =
2
cos 2 B  cos 2 A
cos 2 Bcos 2 A
2
=
sin 2 A  sin 2 B
cos 2 Bcos 2 A
LHS = tan A – tan B
=
sin 2 A
cos 2 A

c.g.road
sin 2 B
cos 2 B
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=
=
=
=
GROUP TUITION
sin 2 A cos 2 B  sin 2 B cos 2 A
cos 2 A cos 2 B
(1  cos 2 A) cos 2 B  (1  cos 2 B) cos 2 A
cos 2 A cos 2 B
cos 2 B  cos 2 A cos 2 B  cos 2 A  cos 2 A cos 2 B
cos 2 A cos 2 B
cos 2 B  cos 2 A
cos 2 B cos 2 A
= RHS(1)
and
=
and
=
and
=
sin 2 A(1  sin 2 B)  sin 2 B(1  sin 2 A)
cos 2 A cos 2 B
sin 2 A  sin 2 A sin 2 B  sin 2 B  sin 2 A sin 2 B
cos 2 A cos 2 B
sin 2 A  sin 2 B
cos 2 Bcos 2 A
= RHS(2)
19. 2(sin6  + cos6 ) – 3(sin4  + cos4 ) + 1 = 0
LHS = 2(sin6  + cos6 ) – 3(sin4  + cos4 ) + 1
= 2(sin2  + cos2 )(sin4  – sin2 cos2 + cos4 ) – 3(sin4  + cos4 ) + 1[ a3 + b3 = (a + b)(a2 – ab + b2)]
= 2(1)(sin4  – sin2 cos2 + cos4 ) – 3(sin4  + cos4 ) + 1
= 2sin4  – 2sin2 cos2 + 2cos4  – 3sin4  – 3cos4  + 1
= 1 – sin4  – 2sin2 cos2 – cos4 
= 1 – (sin4  + 2sin2 cos2 + cos4 )
= 1 – (sin2  + cos2)2
= 1 – (1)2 = 1 – 1 = 0 = RHS
20. If sin  + cos  = p and sec  + cosec  = q then show that q(p2 – 1) = 2p
LHS = q(p2 – 1)
= (sec  + cosec )[(sin  + cos )2 – 1] [ substituting the values of q and p]
= (sec  + cosec )[(sin  + cos )2 – 1]
1
1 
2
2
= 

 [sin  + cos  + 2sin  cos  – 1]
 cos θ sin θ 
sin θ  cos θ 
= 
 [1 + 2sin  cos  – 1]
 cos θ sin θ 
sin θ  cos θ 
= 
 [2sin  cos ]
 cos θ sin θ 
= 2(sin  + cos ) = 2p = RHS
21. If tan  + sin  = a and tan  – sin  = b then prove that a2 – b2 = 4 ab .
tan  + sin  = a and tan  – sin  = b
ab
= (tan  + sin )( tan  – sin )
= tan2  – sin2 
=
sin 2 θ
cos 2 θ
– sin2 
= sin2  
1
 cos 2 θ

1

= sin2 (sec2  – 1)
= sin2 tan2
 ab = sin  tan …(i)
LHS = a2 – b2
= (tan  + sin )2 – (tan  – sin )2
[ substituting the values of a and b]
= (tan  + sin  – tan  + sin )(tan  + sin  + tan  – sin )
= (2sin )(2tan )
= 4 sin  tan 
= 4 ab = RHS [ (i)]
22. a cos  + b sin  = p and a sin  – b cos  = q then prove that a2 + b2 = p2 + q2.
c.g.road
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GROUP TUITION
a cos  + b sin  = p and a sin  – b cos  = q
 p = a cos  + b sin 
and
q = a sin  – b cos 
2
2
 p = (a cos  + b sin )
and
q2 = (a sin  – b cos )2
 p2 = a2cos2  + b2 sin2  + 2abcos  sin 
and
q2 = a2sin2  + b2 cos2  – 2abcos  sin 
RHS = p2 + q2
= a2cos2  + b2 sin2  + 2abcos  sin  + a2sin2  + b2 cos2  – 2abcos  sin 
= a2cos2  + a2sin2  + b2 cos2  + b2 sin2 
= a2(cos2  + sin2 ) + b2(cos2  + sin2 )
= a2(1) + b2(1)
= a2 + b2 = LHS
23. sec  + tan  = p, then obtain the value of sec , tan  and sin  in terms of p.
We know that sec2 – tan2 = 1
(sec  + tan )(sec  – tan ) = 1
But sec  + tan  = p is given
p(sec  – tan ) = 1
sec  – tan  =
1
p
…(i)
adding (i) and (ii)
and
Subtracting (ii) from (i)
1
p
and
2tan = p –
2sec =
p2  1
p
and
2tan =
sec  =
p2 1
2p
2p
and
tan  =
 2sec = p +
cos  =
and
sec  – tan  =
1
p
 sec  + tan  = p
…(ii)
1
p
p 2 1
p
p2 1
2p
p2 1
24. Evaluate the following:
1.
sec38
cosec52
+
2
3
.
tan17 . tan 38 . tan 60 . tan 52 . tan 73 – 3(sin2 32 + sin2 58)
2 .
sec 38
+
tan17 . tan 38 . 3 . cot (90 – 52) . cot (90 – 73) – 3[sin2 32 + cos2 (90 – 58)]
sec( 90  52)
3
sec 38
.
=
+ 2 tan17 . tan 38 . cot 38 . cot 17 – 3[sin2 32 + cos2 32]
sec 38
=
= 1 + 2 . tan17 . cot 38 . tan 38 . cot 17 – 3[1]
= 1 + 2 (tan17 . cot 17)(tan 38 . cot 38) – 3
= 1 + 2 (1)(1) – 3 = 1 + 2 – 3 = 3 – 3 = 0
2.
 cotθ tan(90  θ)  cosecθ sec(90  θ)  sin2 37  sin2 53
tan10 tan20 tan30 tan70 tan80
( cot θ  cotθ  cos ecθ  co sec θ)  (sin 2 37  cos 2 (90  53)
=
1
tan 10  tan 20 
 cot(90  70)  cot(90  80)
3
=
( cot 2 θ  cos ec 2θ)  (sin 2 37  cos 2 (90  53)
1
tan 10  tan 20 
 cot 20  cot 10
3
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GROUP TUITION
(cos ec 2θ  cot 2 θ)  (sin 2 37  cos 2 37)
1
(tan 10  cot 10)(tan 20  cot 20) 
3
11
2
=
=
= 2 3
1
1
(1)(1) 
3
3
=
25. If sin A + cos A = 2 sin(90 – A), then obtain the value of cot A.
sin A + cos A = 2 sin(90 – A)
sin A + cos A = 2 cosA

sin A cos A
2 cos A


sin A sin A
sin A
1 + cotA = 2 cot A
 2 cot A – cot A = 1
cotA( 2 – 1) = 1
1
cot A =
2 1
1
=
2 1
2 1

2 1
2 1
=
( 2 )  (1)
2
26. If cosec  = 2 , then find the value of
2
2 1
=
2 1
=
2 +1
2sin 2θ  3cot 2θ
4tan 2θ  cos 2θ
cosec  = 2
 = 45
sin  = sin 45 =
1
2
, cot  = cot 45 = 1, tan  = tan 45 = 1 and cos  =
1
2
2
 1 
1
  3(1) 2
2
2   3(1)
2
1 3
2 sin θ  3 cot θ

=  2
=  
=
2
1
1


4 tan 2 θ  cos 2 θ
 1 
4
4(1)   

4(1) 2  
2
2
 2
(1  sin θ) (2  2sin θ)
8
If tan  =
then evaluate
.
(2  2cos θ)(1  cos θ)
15
8
15
1
tan  =
 cot  =
=
tan θ
15
8
2
27.

2
=
1 3
4 2
4
8
=
=
=
7
8 1
7
7
2
2
2
(1  sin 2 θ)
225
2(1  sin θ)(1  sin θ)
(1  sin θ)( 2  2 sin θ)
cos 2 θ
15
=
=
=
= cot2  =   =
2
2
(2  2 cos θ)(1  cos θ)
2(1  cos θ)(1  cos θ)
64
(1  cos θ)
sin θ
8
b
28. If cos  =
cos  =
a 2  b2
b
a2  b2
, 0 <  < 90, find the value of sin  and tan .
cos2  =
sin2  = 1 – cos2  = 1 –
sin  =
b2
a  b2
2
b2
=
a2  b2
a2  b2  b2
a2  b2
=
a2
a2  b2
a
a2  b2
a
Also tan  =
sin θ
cos θ
=
a2  b2
b
=
a
b
a2  b2
29. Select a proper option from (a), (b), (c) or (d) from given option so that the statement becomes
correct:
c.g.road
:: 98792 12357 ::
Gurukul
X MATHS gb
23
AJAY PARMAR
GROUP TUITION
1. If  is a measure of an acute angle such that b sin  = a cos , then
(a)
a2  b2
a b
2
a2  b2
a2  b2
(b)
2
(c)
ab
a b
(d)
asinθ  bcosθ
asinθ  bcosθ
=….
a b
ab
b sin  = a cos 
sin  : cos  = a : b Let sin  = ak and cos  = bk

a sin θ  b cos θ
a sin θ  b cos θ
=
k (a 2  b 2 )
a(ak )  b(bk )
a2  b2
= 2 2 = 2 2
a(ak )  b(bk )
a b
k (a  b )
2. Which of the following is correct for some  such that 0   < 90?
1
= 1 (c) sec  = 0
sec θ
1
For  = 0 sec  = 1 
=1
sec θ
(a)
1
>1
sec θ
3. If tan  =
(a)
1
5
2
3
1
tan  =
Now
=
(b)
5
, then
cosec 2θ  sec 2θ
1
3
(b)
(c)
2
3
1
cos θ
<1
=….
(d) 3
cot  = 5
cos ec 2θ  sec 2 θ
cos ec 2θ  sec 2 θ
1  cot 2 θ  1  tan 2 θ
1  cot θ  1  tan θ
2
2
4. If tan2  =
(a)
cosec 2θ  sec 2θ
(d)
7
8
tan2  =
8
7
2  cot θ  tan θ
2
2
, then the value of
(b)
8
7
=
cot 2 θ  tan 2 θ
8
7
cot2  =
(c)
=
 1 

( 5 ) 2  
 5
2
 1 

2  ( 5 ) 2  
 5
(1  sin θ) (1  sin θ)
is …
(1  cos θ) (1  cos θ)
49
64
(d)
49
64
5
2
=
1
5
25

5
=
25  1
35  1
=
24
2
=
3
36
.
7
8
(1  sin θ)(1  sin θ)
cos 2 θ
1  sin 2 θ
7
=
=
= cot2  =
2
2
8
(1  cos θ)(1  cos θ) 1  cos θ
sin θ
cosθ  sinθ
4
5. If cot  = , then the value of
3
cosθ  sinθ
4
4
1
(a) 7
(b)
(c)
(d) –
7
3
3
cos θ sin θ
4

1
cot θ  1
43
cos θ  sin θ
1
= sin θ sin θ =
=3 =
=
cos
θ
sin
θ
4
cos θ  sin θ
7
cot θ  1

1 4  3
sin θ sin θ
3
4
6. If cosec A = and A + B = 90, then sec B = … .
3
1
4
3
7
(a)
(b)
(c)
(d)
4
3
3
3
A + B = 90 B = 90 – A  sec B = sec(90 – A) = cosec A =
4
3
7. If  is the measure of an acute angle and 3 sin  = cos  then  = … .
(a) 30
(b) 45
(c) 60
(d) 90
c.g.road
:: 98792 12357 ::
Gurukul
X MATHS gb
24
AJAY PARMAR
3 sin  = cos  
cos θ
sin θ
GROUP TUITION
= 3 cot  = 3   = 30.
8. If tan A = 5 then find the value of (sin A + cos A) sec A is … .
12
7
12
sin A cos A
1
5
17
(sin A + cos A) sec A = (sin A + cos A) 
=
= tan A + A =
+1=

cos A
12
12
cos A cos A
(a)
12
5
(b)
9. If tan  =
(a)
1
3
7
12
(c)
17
12
(d) –
1  sin θ
=….
1  sin θ
9
3
(c)
(d)
4
16
4
then the value of
3
(b) 3
1  sin θ 1  sin θ
1  sin θ
=
=

1  sin θ 1  sin θ
1  sin θ
(1  sin θ) 2
=
(1  sin θ) 2
=
1  sin θ
cos θ
9  16
5
25
16
But sec2  = 1 + tan2  = 1 +
=
=
sec  =
9
9
9
3
5
4
1
sec  – tan  = – =
3
3
3
2
2
1 sin θ
1
sin θ
=
= sec  – tan 

cosθ cos θ
cosθ
10. In ABC, if mABC = 90, mACB = 45 and AC = 6, then area of ABC = … .
(a) 18
(b) 36
(c) 9
(d)
9
2
mB = 90 and mC = 45  mA = 45
AB = BC = x suppose
Also By Pythagoras theorem AB2 + BC2 = AC2 x2 + x2 = 62 2x2 = 36 x2 = 18
Now Area of ABC = ½  AB  BC = ½  x  x = ½  x2 = ½  18 = 9
11. If cos2 45 – cos2 30 = x cos 45 sin 45 then x = … .
(a) 2
(b)
3
2
(c) –
cos2 45 – cos2 30 = x cos 45 sin 45
2
(d)
3
4
2
 1   3
  x 1  1
  


2
2
2
 2 

 
1
2
1
2
3
4
  
x
2
2 – 3 = 2x 2x = – 1 x = –
1
2
12. If A and B are complementary angles then sin A sec B = … .
(a) 1
(b) 0
(c) – 1
(d) 2
A and B are complementary angles
A + B = 90
A = 90 – B
 sin A sec B = sin (90 – B) sec B = cos B sec B = 1
13. The value of tan 20 tan 25 tan 45 tan 65 tan 70 = … .
(a) – 1
(b) 1
(c) 0
(d) 3
tan 20 tan 25 tan 45 tan 65 tan 70
= tan 20  tan 25  1  cot(90 – 65)  cot(90 – 70)
= tan 20  tan 25  cot 25  cot 20 = (tan 20  cot 20)(tan 25  cot 25) = 1  1 = 1
14. If 7 and 2 are measure of acute angles such that sin 7 = cos 2 then 2sin 3 – 3 tan3 = … .
(a) 1
(b) 0
(c) – 1
(d) 1 – 3
sin 7 = cos 2
 sin 7 = sin (90 – 2)
7 = 90 – 2 9 = 90  = 10 3 = 30
1
2
2sin 3 – 3 tan3 = 2sin 30 – 3 tan 90 = 2( ) – 3 
c.g.road
:: 98792 12357 ::
1
3
=1–1=0
Gurukul
X MATHS gb
25
AJAY PARMAR
15. If A + B = 90, then
GROUP TUITION
cotAcotB  cotAtanB sin 2 B
=….

sinAsecB
cos 2 A
(a) cot2 B
(b) tan2 A
A + B = 90 B = 90 – A
(c) cot2 A
(d) – cot2 A

cotAcotB  cotAtanB sin 2 B

sinAsecB
cos 2 A
=
cotAcot(90  A)  cotAtan(90  A) sin 2 (90  A)

sinAsec(90  A)
cos 2 A
cotAtanA  cotAcotA cos 2 A
=1+ cot2 A – 1 = cot2 A

sinAcosecA
cos 2 A
BC
16. For ABC, sin 
= … .
 2 
A
A
(a) sin
(b) sin A
(c) cos
(d) cos A
2
2
=
In ABC, A + B + C = 180
B + C = 180 – B
BC
= 90
2
BC
 sin 

 2 

A
2
–
A
= sin  90  

A
BC
 sin 
 = cos
2
2


17.
sin4θ  cos 4θ
sin 2θ  cos 2θ
=….
(a) 1
(b) 2
sin θ  cos θ
4
2
4
sin 2θ  cos 2 θ
(c) 3
=
(d) 0
(sin θ  cos θ)(sin θ  cos θ)
2
2
2
2
sin 2θ  cos 2 θ
= sin2  + cos2  = 1
18. If 7cos2  + 3sin2  = 4, then cot  = … .
(a) 7
(b)
7
3
(c) 3
1
(d)
3
7cos  + 3sin  = 4
7cos2  + 3(1 – cos2 ) = 4
7cos2  + 3 – 3cos2 ) = 4
4cos2  = 1
2
2
cos2  =
1
4
cos  =
1
2
 = 60 cot  = cot 30 = 3
19. If tan 5 tan 4 = 1 then  = … .
(a) 7
(b) 3
(c) 10
tan 5 tan 4 = 1
tan 5 =
1
tan4θ
(d) 9
tan 5 = cot 4 tan 5 = tan (90 – 4) 5 = 90 – 4 9 = 90   = 10
20. If A and B are the measures of acute angles and tan A=
(a) 0
tan A =
(b)
1
3
1
2
and sin B =
(c)
1
2
3
2
3
and sin B=
1
2
then cos(A + B) = … .
1
2
 A = 30 and B = 30
 cos(A+B) = cos (30 + 30) = cos 60 =
c.g.road
(d)
1
1
2
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