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Revision Question Bank
Introduction to Trigonometry
1.
Evaluate


sec2 900    cot 2 

2 sin2 250  sin2 650


2cos2 600 tan2 28 tan2 620 cot 400

tan 500
3 sec2 430  cot 2 470


Solution :


sec2 90    cot   cos2 60o tan2 28o tan2 62o cot 40o
+

tan50o
3 sec2 43o  cot 2 47o
2 sin2 25o  sin2 65o




1
2    tan2 28tan2 90o  28o
cosec   cot 
cot 40o
2

=
+
+
o
o
2 sin2 25o  sin2 90o  43o  3 sec2 43o  cot 2 90o  43o  tan  90  40 
2

2







 sec 90o    cosec


1
2
o
2
o
2

tan
28
cot
28
cosec   cot 
cot 40o
4


=
o
2 sin2   cos2 25o 3 sec2 43o  tan2 43o  cot 40
2
2








 sin 90o    cos ,tan 90o    cot  and cot 90o    tan 


1
1
1
2

=
2  1  3  3


1
2
2
2
2
2
2
 1
 cosec   cot   1,sin   cos   1,sec  tan   1 and tan 
cot 


=
2.
1 1
3  1  6 10 5
 1 


2 6
6
6 3
Show that:
1
1

 2sec2  .
1  sin  1  sin 
Solution :
LHS. =
1
1
1  sin   1  sin 


1  sin  1  sin  1  sin  1  sin  
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=
2
1  sin2 
=
2
 2sec2 
2
cos 
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 a2  b2   a  b a  b
 sin2   cos2   1
LHS. = RHS.
3.
Prove the identity (1 + cosec  ) (1 – sin  ) = cos  cot  .
Solution :
To prove, 1  cosec1  sin = cos  cot 
LHS. = 1  cosec 1  sin  
1 

= 1 
 1  sin  
sin  

=
 sin   1 . 1  sin   1  sin 


sin 
sin 
cos2 
=
sin 
=
1 

 cosec 

sin  

 sin2   cos2   1
cos 
.cos 
sin 
cos  

 cot   sin  


= cot .cos 
LHS=RHS
4.
Hence proved
Without using trigonometric tables, evaluate:
cos2 200  cos2 700
× sec2 600 – 2 cot580cot 320 – 4 tan 130 tan 370 tan 450 tan 530 tan 770
sec2 500  cot 2 400
Solution :
cos2 20o  cos2 70o
 2sec2 60o  2cot58o cot32o  4tan13o tan37o tan45o tan53o tan77o
2
o
2
o
sec 50  cot 40

  22  2cot58 cot 90

sec 50  cot  90  50 
1  tan  90  37   tan  90  13 
cos2 20o  cos2 90o  20o
2
o
2
o
o
o
2
o
o
o
o

 58o  4tan13o tan37o
o
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=
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cos 20  sin 20
 8  2cot58o tan58o
2
o
2
o
sec 50  tan 50
2
o
2
o
4tan13o tan37o  cot37o  cot13o




 cos 90o    sin ,ta 90o    cot 


=
1
 8  2 1  4  1  1
1

tan  cot   1
=8–2–4=2
5.
Prove the identity:
1  cos A 1  cos A

 4 cot A cosec .
1  cos A 1  cos A
Solution :
To prove,
1  cos A 1  cos A

 4cot AcosecA
1  cos A 1  cos A
LHS =
1  cos A 1  cos A

1  cos A 1  cos A
1  cos A   1  cos A 
=
1  cos A 1  cos A 
2
=
2
1  cos A  1  cos A 1  cos A  1  cos A 
1  cos A
2
=
22cos A 
sin2 A
=
4cos A 1
.
 4cot A.cosesA
sin A sinA

 a2  b2   a  b a  b
sin2 A  cos2 A  1
cos A
1 

,cosecA 
=  cot A 
sin A
sin A 

LHS =RHS
6.
Hence proved
Show that: (cosec A –1) (cosec A +1) (sec A –1) (sec A +1) =1.
Solution :
LHS
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=  cosecA  1 cosecA  1 sec A  1 sec A  1
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 1
 1
 1
 1

=
 1 
 1 
 1 
 1
 sin A
 sin A
 cos A
 cos A


1 

cosecA

,sec
A


sin A
cos A 

 1  sin A  1  sin A  1  cos A  1  cos A 
=




 sin A  sin A  cos A  cos A 
1  sin A  . 1  cos A 
=
2
2
sin A
=
2
 a2  b2   a  b a  b
2
cos A
cos2 A
sin2 A
.cos
sin2 A
sin2 A
 sin2 A  cos2A  1
=1
LHS=RHS
7.
Show that:
2  cosec2A
sin A  cos A

.
cosec2A  2cot A sin A  cos A
Solution :
1
2  cosec A
sin2 A

LHS =
1
2cos A
cosec2A  2cot A

2
sin A sin A
2
2
1
cos A 

cosecA

and
cot
A


sin A
sin A 

2sin2 A  1
sin2 A
=
1  2sin Acos A
sin2 A
2sin2 A  1
sin2 A

=
Sin2A
1  2sin Acos A
2sin2 A  1
=
sin2 A  cos2 A  2sin Acos A
 sin2 A  cos2 A  1
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
2sin A  sin A  cos A
2
=
=
=
2
2
 sin A  cos A 

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 1  sin2   cos2 
2sin2 A  sin2 A  cos2 A
 sin A  cos A 
2
sin2 A  cos2 A
 sin A  cos A 
2
=
 sin A  cos A  sin A  cos A 
2
 sin A  cos A 
=
sin A  cos A
sin A  cos A
 a2  b2   a  b a  b
LHS=RHS
8.
If 3 cot A = 4 , then check whether
1  tan2 A
 cos2 A  sin2 A or not.
2
1  tan A
Solution :
Let us consider a right angled ABC in which B  90o . For A ,
Base = AB and perpendicular = BC. Also, hypotenuse = AC
3cot A  4
 cot A 
4
3
But cot A =
……..(i)
Base
AB

perpendicular BC
From(i) and (ii) we get
AB 4 4k
 
BC 3 3k
 AB = 4k and BC = 3k
Using Pythagoras theorem
AC2 = AB2 + BC2
 AC2 = (4k)2 + (3k)2
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 AC = 16k 2  9k 2
= 25k 2 
5k 
Now, sin A=
BC 3k 3


AC 5k 5
cos A =
2
 5k
AB 4k 4


AC 5k 5
Also, tan A =
BC 3k 3


AB 4k 4
Now, to check the given equation,
2
3
1  1 9
2
1  tan A
4 
16
LHS 

2
2
1  tan A
 3  1 9
1 
16
4
16  9 7
7
= 16  16 
16  9 25 25
16
16
RHS = cos2 A  sin2 A
2
2
 4   3  16 9 16  9 7



=     
25
25
 5   5  25 25
From Eqs. (i) and (ii),
LHS=RHS
1  tan2 A
 cos2 A  sin2 A

2
1  tan A
9.
Prove the identity (cot A – tan A) cos A = cosec A – 2 sin A.
Solution :
To prove (cot A – tan A) cos A = cosec A – 2sin A
LHS = (cot A – tan A) cos A
 cos A sin A 

=
 cos A
sin
A
cos
A


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cos
2
=
=
A  sin A
2
sin Acos A
 .cos A
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cos2 A  sin2 A
sin A
cos2 A
 sin A
=
sin A
1  sin2 A
=
 sin A
sin A
=
1
 sin A  sin A  cos2   1  sin2 
sin A
= cosec A – 2 sin A
LHS = RHS
Hence proved


1
1
10. Show that cos  

 can be written in the form k tan  and find the
 1  sin  1  sin  
value of k.
Solution :
1
1


cos  


 1  sin  1  sin  
 1  sin   1  sin  
= cos  
 1  sin  1  sin   


 2sin  
= cos  
2 
 1  sin  
= cos .
 a2  b2   a  b a  b
2sin 
sin 
2
2
cos 
cos 
= 2tan
k  2
sin  

tan


cos  

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Chapter Test {Trigonometry}
M: Marks: 40
M: Time: 40 Min.
1. Without using trigonometrical tables, evaluate
cos580 sin 220
cos 380


sin320 cos 680 tan180 tan350 tan600 tan720 tan550
[4]
Solution:
cos 580 sin 220

sin 320 cos 680
cos 380

tan 180 tan350 tan 600 tan 720 tan550
=




cos 900  320 sin 900  680
sin 320

cos 680

cos 380 cosec 900  380




tan180 tan350 3 tan 900  380 tan 900  350

cos 380 sec 380
sin 320 cos 680
=
–

sin 320 cos 680 tan 180 tan 350 3 cot 180 cot350






 cos 900    sin , sin 900    cos , 


cosec 900    sec 



0
and tan 90    cot 




= 11
[
 tan 18 cot18 
0
0
1
tan 350 cot 350

3
cos . sec   1 ]
= 11
=

1
1.1. 3
2 3 1
3
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2. If
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sin  cos 
1
sin  cos 
1
 3 
, then prove that
.


3
3
a
b
a
b
ab
 a  b
4
8
4
8
[4]
Solution:
sin8  cos8 
1
 3 
To prove
3
3
a
b
 a  b
sin4  cos4 
1


Given,
a
b
ab
4
4
 sin  cos  
2
  a  b 

1
b 
 a
 sin4  cos4  
2
2
  a  b 

  sin   cos 
b 
 a
ab 4
ab 4
 
 sin   
 cos 
a
b






2
[
sin2   cos2   1 ]
= sin4   cos 4    sin 4  cos 2 
b
a
sin4   cos4   cos4 
a
b
= sin4   cos4   2sin 2 cos 2 
b
a
 sin4   cos4   2 sin2  cos2   0
a
b
Þ sin4  
2
2
 b
  a

 b
  a

sin2    
cos2   – 2
sin2    
cos2    0
 
 a
  a

 a
  a

2
 b

a
sin2  
cos2    0
 
b
 a



b
a
sin2  
cos2   0
a
b
sin 

cos2 
2
[
a2  b2  2ab   a  b ]
2
[taking square root]
a
b a
b b
a
sin2  cos2

 k  say 

a
b
 sin2  ak and cos2   bk
..(i)
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sin   cos  ak  bk

k
ab
ab
From Eqs. (i) and (ii), we get
sin2  cos2  sin2   cos2 


a
b
ab
sin2  cos2 
1



a
b
ab
Taking first and third terms,
sin2 
1
a

 sin2  
a
ab
ab
Taking second and third terms,
cos2 
1

b
ab
b
 cos2  
ab

2
2

2
sin8  cos8  sin 
Now,
 3 
a3
b
a3
4
 a   ab 
ab ab
=  3   3 
a
b
a
b
=

4
4
 a  b  a  b
=
ab
 a  b
4

..(ii)
.(iii)
..(iv)
  cos 
4
2
4
b3
4
[
from Eqs (iii) and (iv)]
1
 a  b
3
3. If 7 sin2   3cos2   4 and  is an acute angle, then prove that: sec   cosec   2 
2
. [4]
3
Solution:
Given, 7sin2   3cos2   4
 4sin2   3sin2   3 cos2   4


 4sin2   3 sin2   cos2   4
 4sin2   3  4
[
sin2   cos2   1 ]
 4sin2   4  3
 4sin2   1
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1
4
1
 sin  
2
[taking positive square root as  is acute angle]
 sin2  
 cosec   2
sin  
1 
cosec  
cos   1  sin2 
and
1
= 1 
2
= 1
=



2
1

4
4 1
4
3
3

4
2
 sec  



2
3
sec   cosec 
sec  
1 
cos  
2
2
3
Hence proved.
4. Find the acute angles of A and B, if sin (A + 2B) =
3
and cos (A + 4B) = 00, where A > B.
2
[4]
Solution:
Given that, sin (A + 2B) =
3
2
 sin(A + 2B) = sin600
...(i)
 A+ 2B = 60°
and
cos(A + 4B) = 00
 cos (A + 4B) = cos 90°
...(ii)
 A+4B = 90°
On subtracting Eq. (i) from Eq. (ii), we get
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A  4B  90
0
A  2B  600
 

2B  300
 B = 15°
On putting B = 15° in Eq. (i) we get
A + 2(15°) = 60°
A =60° – 30° = 30°
5. If (cot  + tan  ) = m and (sec  – cos  ) = n, then prove that:  mn 
2/3
  mn 
2/3
1.
[4]
Solution:
To prove, (m n)2/3 – (mn )2/3 =1
Given, (cot  + tan  ) = m and (sec  – cos  ) = n
Now, m2n= (cot  tan  )2 (sec  cos  )
2
 cos  sin    1

=

 cos  


 sin  cos    cos 

2
 cos     1  cos2  
=

 
 sin  cos    cos  
1
sin2 
=
.
[ cos2   sin2   1]
2
2
sin  cos  cos 
1
 sec3 
 m2n 
3
cos 
..(i)
and mn2   cot   tan   sec   cos  
2
2
 cos  sin    1

=


cos



 sin  cos    cos 

2
 cos2  sin2   1  cos2  
=


 sin  cos   cos  


sin2 
1

=
sin  cos 
cos2 
2
[
sin2   cos2   1 ]
sin4 
sin3 
=

 mn2  tan3 
3
3
sin  cos  cos 
From Eq. (i), sec3  = m2n
 sec  = (m2n)1/3
…(ii)
[taking cube root both sides] ..(iii)
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tan3 
6.
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From Eq. (ii),
=
[taking cube root both sides] ...(iv)
 sec  = (m2n )1/3
On squaring Eqs. (iii) and (iv) and then subtracting, we get
sec+  – tan2  = (m2n)2/3 – (mn2)2/3
[ sec2  – tan2  = 1]
 1 = (m2n)2/3 –(mn2)2/3
Hence proved.
tan 
cot 

 1  cot   tan  .
Prove that :
1  cot  1  tan 
Solution:
tan 
cot 

To prove,
1  cot  1  tan 
=
tan 
1
1
tan 
LHS =
=
=
mn2
[4]
1
tan 

 1  cot   tan 
1  tan 
tan 
cot 

1  cot  1  tan 
tan 
1
1
tan 
1
tan 

1  tan 



cot  
1 
tan  
tan 
1

tan   1 tan  1  tan  
tan 
tan2 
1
tan3   1
=


tan   1 tan   tan   1 tan   tan   1
 tan   1  tan2   tan   1
=
tan   tan   1
[
a3 – b3 = (a – b) (a2 + ab + b2)]
tan2   tan   1 tan2  tan 
1
=



tan 
tan  tan  tan 
= tan   1  cot   1  cot  tan 
 LHS = RHS
Hence proved.
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7. Prove that :
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1
1
1
1



.
cosec   cot  sin  sin  cosec   cot 
[4]
Solution:
To prove,
1
1
1
1



cosec   cot  sin  sin  cosec   cot 
1
1

LHS =
cosec   cot  sin 
=
=
 cosec   cot   cosec 
1

 cosec   cot   cosec   cot 
 cosec   cot   cosec 
 a  ba  b  a
cosec   cot 
cosec   cot 
 cosec 
=
1
= cosec   cot   cosec 
2



2
[

1
 cosec 
sin 

2
 b2

cosec2  cot 2   1 ]
= cot 
RHS =
1
1

sin  cosec   cot 
= cosec  
= cosec  
= cosec  
 cosec   cot 
1

 cosec   cot   cosec   cot 
 cosec   cot 
cosec
2
  cot 2 
 cosec  cot 
1
= cosec   cosec   cot 

[
cosec2  cot 2   1 ]
= cot 
 LHS = RHS
Hence proved.
1
1
8. If cos   and tan  
then find, Sin      where  and  are both acute angles. [4]
2
3
Solution:
1
1

0
cos
60

Here, cos   cos 60°

2
2 

 a =60°
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and
1
3
tan  



= tan 30°
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1 
3 
tan 300 
   300
 sin      sin 600  300 
= sin 90° = 1
[
Hence, sin (    ) is 1.
sin 90° = 1]
9. Without using trigonometric tables evaluate the following


cosec2 900    tan2  2tan2 300 sec2 520 sin2 380

4 cos2 480  cos2 420
cosec2700  tan2 200

Solution:
Now,
=





cosec2 900    tan2 

4 cos2 480  cos2 420

4[{cos  90  42  }  cos
[4]

2tan2 300 sec2 520 sin2 380

cosec2 700  tan2 200



[{cosec 900   }2  tan2  ]
0
2
0
2
2
420 ]
 1 
0
0 2
2
0
2 
 . [sec 90  38 } . sin 38
 3

{cosec2 900  200 }2  tan2 200




2
.cosec2380. sin 2 38 0
sec   tan 
3 2 0
=
2
0
2
0
4{sin 42  cos 42 }
sec 20  tan2 200
2
2






 cosec 900    sec  


0

cos 90    sec  


and sec 900    cosec 


=
1 2
1
 . 2 0 . sin2 380
4 3 sin 38


 sec2   tan2   1 


2
2
 sin   cos   1 

1 
and cosec  

sin  

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1 2 1
 .
4 3 1
3  8 5

=
12 12
=
10. Prove that:
1  cot A  tan A   sin A  cos A  = sin 2 A cos 2 A.
sec
3
A  cosec3A
Solution:
To prove,
1  cot  tan A  sin A  cos A 
sec
3
LHS =
A  cosec A
3


[4]
 sin2 Acos2 A
1  cot A  tan A  sin A  cos A 
sec
3
A  cosec3A

cos A sin A 

 1  sin A  cos A   sin A  cos A 

=
1 
 1

 cos3 A sin3 A 


sin A
cos A


tan
A

,
cot
A



cos A
sin A


and sec A  1 , cosec A  1 

cos A
sin A 

cos2 A  sin2 A 
1 
  sin A  cos A 
sin A cos A 

=
sin3 A  cos3 A
sin3 A  cos3 A


1
 1  sin A cos A   sin A  cos A 

=
3
sin A  cos3 A
sin3 Acos3 A
3
3
 sin A cos A  1  sin Acos A   sin A  cos A 
[sin   cos   1]  

sin3 A  cos3 A
 sin A cos A 
2
2
 sin A cos A  1
=
sin A cos A
[
sin
3
×

A cos3 A   sin A  cos A 
 sin A  cos A   sin2 A  cos2 A  sin Acos A 
a3 – b3 = (a – b) (a2 + b2 + ab)]
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=
=
 sin A cos A  1  sin
2
2
Acos A
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
sin2 A  cos2 A  sin A cos A
 sin A cos A  1  sin2 Acos2 A 
1  sin A cos A
= sin2 A cos2 A
 LHS = RHS
[
sin2A + cos2 A = 1]
Hence proved.
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