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Revision Question Bank
Introduction to Trigonometry
1. Evaluate
2 0 2 2 0 2 2 0 0
02 0 2 0 2 0 2 0
sec 90 cot 2cos 60 tan 28 tan 62 cot 40
tan 502 sin 25 sin 65 3 sec 43 cot 47
Solution :
2
2 o 2 o
sec 90 cot
2 sin 25 sin 65
+
2 o 2 o 2 o o
o2 o 2 o
cos 60 tan 28 tan 62 cot 40
tan503 sec 43 cot 47
=
2 2
2 o 2 o o
cosec cot
2 sin 25 sin 90 43
+
2 2 o o
2 o 2 o o
12 tan 28tan 90 28
2
3 sec 43 cot 90 43
+
o
o o
cot 40
tan 90 40
osec 90 cosec
=
2 o 2 o2 2 o
o2 o 2 o2 2 o
12 tan 28 cot 28cosec cot cot 404
cot 403 sec 43 tan 432 sin cos 25
o o osin 90 cos ,tan 90 cot and cot 90 tan
=
111 2
2 1 3 3
2 2 2 2 2 2 1cosec cot 1,sin cos 1,sec tan 1 and tan 1
cot
= 1 1 3 1 6 10 5
12 6 6 6 3
2. Show that: 21 12sec
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 2
2
2a b a b a b
1 sin
= 2
2
22sec
cos
2 2sin cos 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 1 sin cosecsin sin
=
sin 1 1 sin
. 1 sinsin sin
= 2cos
sin
2 2sin cos 1
= cos
.cossin
= cos
cot .cos cotsin
LHS=RHS Hence proved
4. Without using trigonometric tables, evaluate:
2 0 2 0
2 0 2 0
cos 20 cos 70
sec 50 cot 40
× sec2 600 – 2 cot580cot 320 – 4 tan 130 tan 370 tan 450 tan 530 tan 770
Solution :
2 o 2 o2 o o o o o o o o
2 o 2 o
cos 20 cos 702sec 60 2cot58 cot32 4tan13 tan37 tan45 tan53 tan77
sec 50 cot 40
2 o 2 o o2 o o o o o
2 o 2 o o
o o o o
cos 20 cos 90 202 2 2cot58 cot 90 58 4tan13 tan37
sec 50 cot 90 50
1 tan 90 37 tan 90 13
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= 2 o 2 o
o o
2 o 2 o
cos 20 sin 208 2cot58 tan58
sec 50 tan 50
o o o o4tan13 tan37 cot37 cot13
o ocos 90 sin ,ta 90 cot
= 1
8 2 1 4 1 1 tan cot 11
= 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 cosA 1 cosA4cot AcosecA
1 cosA 1 cosA
LHS = 1 cosA 1 cosA
1 cosA 1 cosA
=
2 21 cos A 1 cos A
1 cos A 1 cos A
=
2
1 cos A 1 cos A 1 cos A 1 cos A
1 cos A
2 2a b a b a b
= 2 2
2
2 2cosAsin A cos A 1
sin A
= 4cosA 1
. 4cot A.cosesAsin A sinA
= cos A 1
cot A ,cosecAsin A sin A
LHS =RHS Hence proved
6. 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 secA 1 secA 1
= 1 1 1 1
1 1 1 1sin A sin A cos A cos A
1cosecA ,sec A
sin A cosA
= 1 sin A 1 sin A 1 cos A 1 cos A
sin A sin A cos A cos A
= 2 2
2 2
1 sin A 1 cos A.
sin A cos A
2 2a b a b a b
=2 2
2 2
2 2
cos A sin A.cos sin A cos A 1
sin A sin A
= 1
LHS=RHS
7. Show that: 2
2
2 cosec A sin A cosA
cosec A 2cot A sin A cosA
.
Solution :
LHS = 2 2
2
2
12
2 cosec A sin A1 2cos Acosec A 2cot A
sin A sin A
1 cosAcosecA and cot A
sin A sin A
=
2
2
2
2sin A 1
sin A1 2sin Acos A
sin A
= 2 2
2
2sin A 1 sin A
Sin A 1 2sin Acos A
= 2
2 2
2sin A 1
sin A cos A 2sin Acos A
2 2sin A cos A 1
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=
2 2 22sin A sin A cos A
sin A cos A
2 21 sin cos
=
2 2 2
2
2sin A sin A cos A
sin A cos A
=
2 2
2
sin A cos A
sin A cos A
=
2
sin A cos A sin A cos A
sin A cos A
2 2a b a b a b
= sin A cos A
sin A cos A
LHS=RHS
8. If 3 cot A = 4 , then check whether
22 2
2
1 tan Acos A sin A or not.
1 tan A
Solution :
Let us consider a right angled ABC in which oB 90 . For A ,
Base = AB and perpendicular = BC. Also, hypotenuse = AC
3cot A 4
4cot A
3 ……..(i)
But cot A = 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 = 2 216k 9k
= 2225k 5k 5k
Now, sin A= BC 3k 3
AC 5k 5
cos A = AB 4k 4
AC 5k 5
Also, tan A = BC 3k 3
AB 4k 4
Now, to check the given equation,
2
2
22
3 91 11 tan A 4 16LHS
91 tan A 3 11 164
=
16 9 7716 16
16 9 25 2516 16
RHS = 2 2cos A sin A
= 2 2
4 3 16 9 16 9 7
5 5 25 25 25 25
From Eqs. (i) and (ii),
LHS=RHS
2
2 2
2
1 tan Acos A sin A
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 Asin A cos A
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= 2 2cos A sin A
.cos Asin Acos A
= 2 2cos A sin A
sin A
= 2cos A
sin Asin A
= 21 sin A
sin Asin A
= 2 21sin A sin A cos 1 sin
sin A
= cosec A – 2 sin A
LHS = RHS Hence proved
10. Show that cos 1 1
1 sin 1 sin
can be written in the form k tan and find the
value of k.
Solution :
1 1cos
1 sin 1 sin
= 1 sin 1 sin
cos1 sin 1 sin
= 2
2sincos
1 sin
2 2a b a b a b
= 2
2sin sincos . 2
cos cos
= 2tan
k 2 sin
tancos
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Chapter Test {Trigonometry}
M: Marks: 40 M: Time: 40 Min.
1. Without using trigonometrical tables, evaluate 0 0 0
0 0 0 0 0 0 0
cos58 sin 22 cos 38
sin32 cos 68 tan18 tan35 tan60 tan72 tan55
[4]
Solution: 0 0
0 0
cos 58 sin 22
sin 32 cos 68
0
0 0 0 0 0
cos 38
tan 18 tan35 tan 60 tan 72 tan55
= 0 0 0 0
0 0
cos 90 32 sin 90 68
sin 32 cos 68
0 0 0
0 0 0 0 0 0
cos 38 cosec 90 38
tan18 tan35 3 tan 90 38 tan 90 35
= 0 0
0 0
sin 32 cos 68
sin 32 cos 68 –
0 0
0 0 0 0
cos 38 sec 38
tan 18 tan 35 3 cot 18 cot35
0 0
0
0
cos 90 sin , sin 90 cos ,
cosec 90 sec
and tan 90 cot
= 0 0 0 0
11 1
tan 18 cot18 tan 35 cot 35 3
[ cos . sec 1 ]
= 1
1 11.1. 3
= 2 3 1
3
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2. If 4 4sin cos 1
a b a b
, then prove that
8 8
33 3
sin cos 1
a b a b
. [4]
Solution:
To prove
8 8
33 3
sin cos 1
a b a b
Given, 4 4sin cos 1
a b a b
4 4
2sin cosa b 1
a b
4 4
22 2sin cos
a b sin cosa b
[ 2 2sin cos 1 ]
4 4a b a bsin cos
a b
= 4 4 4 2sin cos sin cos
Þ 4 4 4 4b asin sin cos cos
a b
= 4 4 2 2sin cos 2sin cos
4 4 2 2b asin cos 2 sin cos 0
a b
2 2
2 2b asin cos
a a
– 2 2b a
2 sin cos 0a a
2
2 2b asin cos 0
a b
[
22 2a b 2ab a b ]
2 2b asin cos 0
a b [taking square root]
2
2
asin abcos bb
a
2 2sin cos
k saya b
..(i)
2 2sin ak and cos bk
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2 2sin cos ak bk
ka b a b
..(ii)
From Eqs. (i) and (ii), we get 2 2 2 2sin cos sin cos
a b a b
2 2sin cos 1
a b a b
Taking first and third terms, 2
2sin 1 asin
a a b a b
.(iii)
Taking second and third terms, 2cos 1
b a b
2 bcos
a b
..(iv)
Now,
4 42 28 8
3 3 3 3
sin cossin cos
a b a b
=
4 4
3 3
a ab
a b a b
a b
[ from Eqs (iii) and (iv)]
=
4 4
a b
a b a b
=
4 3
a b 1
a b a b
3. If 7 2 2sin 3cos 4 and is an acute angle, then prove that:2
sec cosec 2 .3
[4]
Solution:
Given, 2 27sin 3cos 4
2 2 24sin 3sin 3 cos 4
2 2 24sin 3 sin cos 4
24sin 3 4 [ 2 2sin cos 1 ]
24sin 4 3
24sin 1
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2 1sin
4
1
sin2
[taking positive square root as is acute angle]
cosec 2 1
sincosec
and 2cos 1 sin
= 2
11
2
= 1 4 1
14 4
= 3 3
4 2
2
sec3
1
seccos
2sec cosec 2
3
Hence proved.
4. Find the acute angles of A and B, if sin (A + 2B) = 3
2 and cos (A + 4B) = 00, where A > B.
[4]
Solution:
Given that, sin (A + 2B) = 3
2
sin(A + 2B) = sin600
A+ 2B = 60° ...(i)
and cos(A + 4B) = 00
cos (A + 4B) = cos 90°
A+4B = 90° ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
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0
0
0
A 4B 90
A 2B 60
2B 30
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: 2/3 2/3
mn mn 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 )
=
2cos sin 1
cossin cos cos
=
2 2cos 1 cos
sin cos cos
= 2
2 2
2 2
1 sin. [ cos sin 1]
sin cos cos
2 3
3
1m n sec
cos
..(i)
and 22mn cot tan sec cos
=
2cos sin 1
cossin cos cos
=
22 2 2cos sin 1 cos
sin cos cos
=
22
2
sin1
sin cos cos
[ 2 2sin cos 1 ]
= 4 3
2 3
3 3
sin sinmn tan
sin cos cos
…(ii)
From Eq. (i), sec3 = m2n
sec = (m2n)1/3 [taking cube root both sides] ..(iii)
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From Eq. (ii), tan3 = mn2
sec = (m2n )1/3 [taking cube root both sides] ...(iv)
On squaring Eqs. (iii) and (iv) and then subtracting, we get
sec+ – tan2 = (m2n)2/3 – (mn2)2/3
1 = (m2n)2/3 –(mn2)2/3 [ sec2 – tan2 = 1]
Hence proved.
6. Prove that :tan cot
1 cot tan1 cot 1 tan
. [4]
Solution:
To prove, tan cot
1 cot 1 tan
=
1
tan tan1 cot tan
1 1 tan1tan
LHS = tan cot
1 cot 1 tan
=
1
tan tan1 1 tan1
tan
1
cottan
=
tan 1
tan 1 tan 1 tantan
=
2 3tan 1 tan 1
tan 1 tan tan 1 tan tan 1
=
2tan 1 tan tan 1
tan tan 1
[ a3 – b3 = (a – b) (a2 + ab + b2)]
= 2 2tan tan 1 tan tan 1
tan tan tan tan
= tan 1 cot 1 cot tan
LHS = RHS Hence proved.
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7. Prove that :1 1 1 1
.cosec cot sin sin cosec cot
[4]
Solution:
To prove,
1 1 1 1
cosec cot sin sin cosec cot
LHS = 1 1
cosec cot sin
=
cosec cot1cosec
cosec cot cosec cot
1cosec
sin
=
2 2
cosec cotcosec
cosec cot
2 2a b a b a b
= cosec cot
cosec1
[ 2 2cosec cot 1 ]
= cosec cot cosec
= cot
RHS = 1 1
sin cosec cot
=
cosec cot1cosec
cosec cot cosec cot
=
2 2
cosec cotcosec
cosec cot
= cosec cot
cosec1
[ 2 2cosec cot 1 ]
= cosec cosec cot
= cot
LHS = RHS Hence proved.
8. If 1 1
cos and tan2 3
then find, Sin where and are both acute angles. [4]
Solution:
Here, cos 1
2 cos 60° 0 1
cos 602
a =60°
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and 1
tan3
= tan 30° 0 1tan 30
3
030
0 0sin sin 60 30
= sin 90° = 1 [ sin 90° = 1]
Hence, sin ( ) is 1.
9. Without using trigonometric tables evaluate the following
2 0 2 2 0 2 0 2 0
2 0 2 0 2 0 2 0
cosec 90 tan 2tan 30 sec 52 sin 38
4 cos 48 cos 42 cosec 70 tan 20
[4]
Solution:
Now,
2 0 2 2 0 2 0 2 0
2 0 2 0 2 0 2 0
cosec 90 tan 2tan 30 sec 52 sin 38
4 cos 48 cos 42 cosec 70 tan 20
=
0 2 2
0 0 2 2 0
[{cosec 90 } tan ]
4[{cos 90 42 } cos 42 ]
2
0 0 2 2 0
2 0 0 2 2 0
12 . [sec 90 38 } . sin 38
3
{cosec 90 20 } tan 20
=
2 0 2 02 2
2 0 2 0 2 0 2 0
2.cosec 38 . sin 38
sec tan 34{sin 42 cos 42 } sec 20 tan 20
0
0
0
cosec 90 sec
cos 90 sec
and sec 90 cosec
= 2 0
2 0
1 2 1. . sin 38
4 3 sin 38
2 2
2 2
sec tan 1
sin cos 1
1and cosec
sin
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= 1 2 1
.4 3 1
= 3 8 5
12 12
10. Prove that:
3 3
1 cot A tan A sin A cos A
sec A cosec A
= sin2A cos2A. [4]
Solution:
To prove,
2 2
3 3
1 cot tan A sin A cos Asin Acos A
sec A cosec A
LHS =
3 3
1 cot A tan A sin A cos A
sec A cosec A
=
3 3
cos A sin A1 sin A cos A
sin A cos A
1 1
cos A sin A
sin A cos Atan A , cot A
cos A sin A
1 1and sec A , cosec A
cos A sin A
=
2 2
3 3
3 3
cos A sin A1 sin A cos A
sin A cos A
sin A cos A
sin A cos A
=
3 3
3 3
11 sin A cos A
sin A cos A
sin A cos A
sin Acos A
3 3
2 2
3 3
sin Acos A sin A cos Asin A cosA 1[sin cos 1]
sin A cos A sin A cos A
= sin A cos A 1
sin A cos A
×
3 3
2 2
sin A cos A sin A cos A
sin A cos A sin A cos A sin Acos A
[ a3 – b3 = (a – b) (a2 + b2 + ab)]
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= 2 2
2 2
sin A cos A 1 sin Acos A
sin A cos A sin A cos A
= 2 2sin A cos A 1 sin Acos A
1 sin A cos A
[ sin2A + cos2 A = 1]
= sin2 A cos2 A
LHS = RHS Hence proved.