ML Aggarwal Solution Class 9 Chapter 17 Trigonometric Ratios Exercise 17.1

 Exercise 17.1

Question 1

Sol :
(a)







By Pythagoras theorem
OP2=OM2+PM2
152=122+PM2
225=144+PM2
PM2=225144
PM2=81
PM=81
PM=9 

(i) sinθ=PMOP
=915
sinθ=35

(ii) cosθ=OMOP
cosθ=1215
cosθ=45

(iii) tanθ=PMOM
=912
tanθ=34

(iv) cotθ=OMPM
=129
cotθ=43

(v) secθ=OPOM
=1512
secθ=54

(vi) cosecθ=OPPM
=159
cosecθ=53









(b) By Pythagoras theorem 
AB2=AC2+BC2
=122+52
=144+25
AB2=169
AB=169
AB=13

(i) Sin A=BCAB
Sin A=513

(ii) Cos A=ACAB
Cos A=1213

(iii) sin2A+cos2A=(513)2+(1213)2
=25169+144169
=25+144169
=169169
=1
sin2A+cos2A=1


(iv) sec2Atan2A
secA=1cosA
=11213

tanA=sinAcosA
=(513)(1213)

TanA=512
sec2Atan2A=(1312)2(512)2
=16914425144
=16925144
=144144
=1
=sec2Atan2A=1


Question 2

Sol :
(a) 







By Pythagoras theorem
hypotenuse =BC
BC2=AB2+AC2
102=62+AC2
100=36+AC2
AC2=10036
AC2=64
AC=64
AC=8

(i) Sin B=ACBC
=810

Sin B=45

(ii) Cos C=ACBC
=810

cosC=45

(iii) Sin B+ Sin C
sinC=ABBC
=610
=35
sinB=45
sinB+sinC=45+35
=4+35

Sin B + Sin C=75

(iv) Sin B. Cos C+ Sin C. cos B
Cos B=ABBC=610
=35

sinBcosC+sinCcosB
4545+3535
1625+925
16+925
2525
⇒1











(b) From Figure ΔADC
AD2+CD2=AC2
AD2+52=132
AD2+25=169
AD2=16925
AD2=144
AD=√144=12

From figure ΔABD
⇒BD=BC-CD
=21-5
BD=16

AB2=BD2+AD2
AB2=162+122
AB2=256+144
AB2=400
AB=400
AB=20

∴AB=20, AD=12
BD=16
AC=13
CD=5

(i) tan x=CDAD
=512

(ii) cos y =BDAB
=1620

Cos y=45

(iii) cosec2ycot2y

cosecy=ABAD
=2012

coty=BDAD
=1612
coty=43

cosec2ycot2y=(53)2(43)2
=259169
=25169
=99

cosec2ycot2y=1

Question 3

Sol :









From figure
BD=BC-CD
BD=21-5=16

From ΔADC
AD2+DC2=AC2
AD2+52=132
AD2+25=169
AD2=16925
AD2=144
AD=144
AD=12

Secθ=ABBD
From figure ΔABD
AB2=BD2+AD2
=162+122
=256+144=400

AB2=400
AB=400
AB=20
secθ=2016
=54










(b) From figure ΔABC
AC2=AB2+BC2
=32+42
=9+16
AC2=2b
AC=25
AC=5

From figure ΔBCD
CD2=BC2+BD2
122=42+BD2

144=16+BD2
BD2=14416
=128

BD2=128
BD=128
=16×4×2
=4×22
BD=82

(i) Sin x=BCAC
sinx=45

(ii) cotx=ABBC
=34

(iii) cot2xcosec2x
cotx=34
cosecx=ACBC
=54

cot2xcosec2x=(34)2(54)2
=916254

cot2xcosec2x=92516
=1616
=1

(iv) secy=CDBD
=1282

secy =322


(v) tan2y1cos2y
sec2y=1cos2y
=(322)2
=94×2

1cos2y=98

Tan2y=(BCBD)2
=(482)2
=1664×2

Tan2y=18

tan2y1cos2y=1898
=198
=88
=-1

Question 4

Sol :
(a) 








From figure ΔBCD
BC2=BD2+CD2
B2=92+122
=81+144
BC2=225
BC=225
BC=15

ΔABC⇒AB2+BC2=AC2
AB2+152=252
AB2=625225
AB2=400
AB=400=20

(i) 2sin y-cos y
siny=BDBC
=915

Siny =35
cosy=CDBC
=1215
cosy=45
2sinycosy=23545
=6545
=645=25


(ii) 2sin x-cos x
sinx=BCAC=1525=35
cosx=ABAC=2025=45
2sinxcosx=23545
=6545
=645=25

(iii) 1-sin x+cos y
sinx=35
cosy=45

1sinx+cosy
=135+45
=53+45
65

(iv) 2cosx-3siny+4tanx 
sinx=35
cosx=45
siny=35

tanx=sinxcosx
=3545
tanx=34

∴2cosx-3siny+4tanx

245335+434

8595+3
89+155
145










(b) By Pythagoras theorem
(ii) AC2=BC2+AB2
52=32+AB2
25=9+AB2
AB2=259
AB2=16
AB=16
AB=4
∴AB=y=4

(i) Sinx =BCAC

sinx=35


Question 5

Sol :






Given : TanA=yzxy=512
By Pythagoras theorem

xy2+yz2=xz2
122+52=xz2
∴144+25=xz2
xz2=169
xz=13

(i) cos A=xyxz

=1213


(ii) cosec A-cot A
cosecA=xzyz
=135

cot A=xyyz
=125

∴cosec A-cot A=135125
=13125=15

Question 6








Sol :
(a) Given : AB=7
BC-AC=1

By Pythagoras theorem
BC2=AB2+AC2
∵BC=1+AC
(1+AC)2=72+AC2
1+AC2+2AC=49+AC2
2AC=49-1
2AC=48
AC=24

(i) Sin C=ABBC
∵From figure 
BC-AC=1
BC-24=1
BC=1+24
BC=25

Sin C=725

(i) tan B=ACAB
=247


(b)








Given PQ=40
PR+QR=50

By Pythagoras theorem
PR2=PQ2+QR2
(50QR)2=PQ2+QR2
2500+QR2100QR=402+QR2
2500-1600=100QR
100QR=900
QR=900100
QR=9

∴Given 
PR+QR=50
PR+9=50
PR=50-9=41

(i) Sin P=QRPR
=941

(ii) Cos P =PQPR
=4041

(iii) tan R=sinRcosR.
SinR=PQPR
=4041

Cos R=QRPR
=941

∴tan R =4041941
=409

∴tan R=409

Question 7

Sol :










(i)
cos < ABC 

=BDBA

=915=35



(ii)
sin < ACB=ADAC
=1215=45



Question 8

Sol :
(a) 







Given : AB=AC=5 cm
BC=6

(i) sin C=ABAC
=45

(ii) tan B=ADBD
=43

(iii) tan C-cot B
∵tan C =ADDC=43
cot B =BDAD=34

⇒tan C-cot B=4334
=16912=712


(b) 








Given AB=2
BC=1

∴sin θ=ABAC
tan θ=ABBC

∴By Pythagoras theorem
AC2=AB2+BC2
AC2=22+12
AC2=4+1
AC2=5
AC=5

sin2θ=(ABAC)2=(25)2=45
tan2θ=(ABBC)2=(21)2=4

sin2θ+tan2θ=45+4
=4+205
=245
sin2θ+tan2θ=445

(c) 









Given BD=15
Sin B=45
tan C=1

sinB=45×55
SinB=ADAB=2025

(i) AD=20
tanCsinCcosC=1
ADDC=1
∴AD=DC
∴DC=20

In ΔACD
By Pythagoras theroem

AC2=AD2+DC2
=202+202
=400+400

AC2=800
AC=800
=400×2
AC=202

(ii) tan2B1cos2B=1

LHStan2B1cos2B

tan2B=(ADBD)2=(.2015)2=(43)2
=169

cos2B=(BDAB)2=(1525)2=(35)2=925


tan2B1cos2B1691(925)
169259
16259
1
∴LHS=RHS

Hence proved


Question 9

Sol :






Given : sinθ=35

(i) ∴From Pythagoras theorem
bc2=ab2+ac2
52=32+ac2
25=9+ac2
ac2=259
ac2=16
ac=16
ac=4

cosθ=acbc
=45

(ii) tanθ=sinθcosθ
=3848
=34

Question 10

Sol :






Given that : tanθ=512
By Pythagoras theorem
bc2=ab2+ac2
=52+122
=25+144
bc2=169
bc=169
∴bc=13

sinθ=abbc
=513

cosθ=acbc
=1213


Question 11

Sol :







Given : sinθ=610

By Pythagoras theorem
bc2=ab2+ac2
102=62+ac2
100=36+ac2
ac2=10036
ac2=64
ac=64
ac=8

cosθ=acbc=810

tanθ=abac=68

cosθ+tanθ=810+68
=64+6080
=124803120

Question 12

Sol :





By Pythagoras theorem
ab2+ac2=bc2
42+32=bc2
16+9=bc2
bc2=25
bc=25
bc=5

sinθ=abbc=45
cosθ=acbc=35

sinθ+cosθ=45+35
=4+35
=75


Question 13








Sol :
Given : cosecθ=√5
cosecθ=1sinθ=bcab=51
bc=5 ; ab=1

By Pythagoras theorem
bc2=ab2+ac2
(5)2=12+ac2
5=1+ac2
ac2=51
ac2=4
ac=4
ac=2

cotθcosθ=acabacbc
=2125
=2525
=2(51)5

Question 14

Sol :






Given : sinθ=pq
By Pythagoras theorem
q2=p2+ac2
ac2=a2p2
ac=q2p2

cos θ=acbc=p2+q2q
∴cos θ +sin θ=q2p2q+pq
=p+q2p2q

Question 15






Sol :
Given tanθ=815
By Pythagoras theorem

bc2=ab2+ac2
bc2=82+152
bc2=64+225
bc2=289
bc=289
bc=17

∴ab=8 ; ac=15 ; bc=17

sec θ=bcac=1715
cosec θ=bcab=178

∴secθ+cosecθ=1715+178
=17×8+17×15120
=391120
=331120

Question 16







Sol :
Given : 
13 sin A=5
sinA=513

By Pythagoras theorem
BA2=CB2+AC2
132=52+AC2
169=25+AC2
AC2=16925
AC2=144
AC=144
AC=12

∴CB=5 ; BA=13 ; AC=12

sinA=513

cosA=1213

tanA=sinAcosA
=(513)(1213)
512

5sinA2cosAtanA=5(513)2(1213)(512)

=25132413512
=252413512
=113×125=1265

Question 17







Sol :
Given : cosec A=√2

cosecA=1sinA=2
sinA=12

By Pythagoras theorem

bc2=ab2+ac2

(2)2=12+ac2

2=1+ac2

ac2=21

ac2=1

ac=1

ac=1


∴ac=1 ; bc=√2 ; ab=1


∴sin A=12


cos A=12


tan A=sinAcosA
=(12)(12)=1

cot A1tanA
=11

cot A=1

2sin2A+3cot2Atan2Acos2A

2(12)2+3(1)2(1)2(12)2

212+3(1)112
1+3(212)
4×21
⇒8

Question 18

Sol :









Given : ABCD is a rhombus
AC=8 ; BD=6
From figure ΔOBC







By Pythagoras theorem
BC2=OC2+OB2
=32+42
=9+16=25
BC=25=5


sin < OCB=OBBC
=35


Question 19

Sol :






Given : tanθ=512

By Pythagoras theorem
bc2=ab2+ac2
bc2=52+122
bc2=25+144
bc2=169
bc=109
bc=13

∴bc=13 ; ab=5 ; ac=12

sinθ=abbc=513

cosθ=acbc=1213

cosθ+sinθcosθsinθ=1213+5131213513

=12+51312513

=1778=237

Question 20

Sol :







Given : 5cos A-12 sin A=0
5cos A=12sin A
512=sinAcosA
512=tanA

By Pythagoras theorem
bc2=ab2+ac2
bc2=52+122
bc2=25+144
bc2=169
bc=169
bc=13

∴ab=5 ; bc=13 ; ac=12

Sin A=abbc=513

cos A=acbc=1213

sinA+cosA2cosAsinA
=513+121321213513
=(5+12)/13(245)/13
=1719

Question 21







Sol :
Given : tanθ=pq
By Pythagoras theorem
bc2=ab2+ac2
bc2=p2+q2
bc=p2+q2
∴sin θ=abbc=pp2+q2

cos θ=acbc=qp2+q2

psinθqcosθpsinθ+qcosθ
=ppp2+q2qqp2+q2ppp2+q2+qqp2+q2

=p2q2p2+q2p2+q2p2+q2

p2a2p2+q2

Question 22








Sol :
Given : 3cotθ=4
cotθ=43

By Pythagoras theorem
bc2=ab2+ac2
bc2=32+42
bc2=9+16
bc2=25
bc=25
bc=5

sinθ=abbc=35

cosθ=acbc=45

5sinθ3cosθ5sinθ+3cosθ
=5.35345535+345
=(1512)/5(15+12)/5
=327=19

Question 23

Sol :
(i)






Given : 5cosθ-12sinθ=0
⇒5cosθ=12sinθ
512=sinθcosθ
512=tanθ

By Pythagoras theorem
bc2=ab2+ac2
bc2=52+122
bc2=25+144
bc2=169
bc=169
bc=13

∴sinθ=abbc=513

cosθ=acbc=1213
sinθ+cosθ2cosθsinθ
=513+121321213513
=5+121324513
=1716

(ii)






Given : cosecθ=1312
cosecθ=1sinθ=1312
sinθ=1213

By Pythagoras theorem
bc2=ab2+ac2
132=122+ac2
169=144+ac2
169144=ac2
25=ac2
ac=25
ac=5

∴sinθ=1213

cosθ=acbc=513

2sinθ3cosθ4sinθ9cosθ
=212133513412139513
=241513484513
=93
=3

Question 24








Sol :
Given : 
5sinθ=3
sinθ=35

By Pythagoras theorem
bc2=ab2+ac2
52=32+ac2
25=a+ac2
ac2=259
ac2=16
ac=16
ac=4

cosθ=acbc=45

∴secθ=1cosθ=54

tanθ=sinθcosθ=3/54/5=34

secθtanθsecθ+tanθ
=543454+34

=5345+34=28=14

Question 25

Sol :
Given :
sinθ=cosθ
Then 2tan2θ+sin2θ1
2(sinθcosθ)2+sin2θ1
2(sinθsinθ)2+sin2θ1
2(1)2+sin2θ1
2+sin2θ1
sin2θ+1

From given sinθ=cosθ
So, θ=45
∴sin45=cos45
12=12

2tan2θ+sin2θ1
sin2θ+1
(12)2+1
12+1=32

2tan2θ+sin2θ1=32

Question 26

Sol :
(i)
L.H.S⇒cosθ.tanθ
⇒cosθsinθcosθ (tanθ=sinθcosθ)
⇒sinθ

R.H.S=sinθ
∴L.H.S=R.H.S

(ii)
LHS⇒sinθ.cotθ
sinθcosθsinθ [cotθ=cosθsinθ]
⇒cosθ
∴L.H.S=R.H.S

(iii)
LHS⇒sin2θcosθ+cos2θ
sin2θ+cosθcosθcosθ
sin2θ+cos2θcosθ (sin2θ+cos2θ=1)
1cosθ
∴L.H.S=R.H.S

Question 27








Sol :
Given : ∠C=90°

tanA=34
By Pythagoras theorem
AB2=BC2+AC2
AB2=32+42
AB2=9+16
AB2=25
AB=25
AB=5

sinA=BCAB=35

cosA=ACAB=45

cosB=BCAB=35

SinB=ACAB=45


∴sinA cosB+cosAsinB
3535+4545
925+1625
9+1625=2525
⇒1
∴L.H.S=R.H.S

Question 28

Sol :
(a)










Given : 
In ΔABC, ΔBRS
AB=18 cm
BC=7.5 cm
RS=5 cm
∠BSR=x°
∠SAB=y°

From Hint ; AR=12 cm
RB=6 cm
AC=19.5 cm

(i) tanx=RBSR=65
(ii) siny=BCAC
=7.519.5×1010
=75195=513

∴sin y=513

(b)








By Pythagoras theorem
AC2=AB2+BC2
AC2=122+52
AC2=144+2
AC2=169
AC=169
AC=13

∴cos < CBD=ABAC
=1213

(ii) cot<ABD=BCAC
=513

Question 29









Sol :
Given :
ABCD is rectangle
AC=15
∠ACD=ɑ ; cot ɑ=32
cotα=CDAD=32
CD=32AD

From ΔACD

AC2=AD2+CD2
152=AD2+(32AD)2
225=AD2+9AD24
225=4AD2+9AD24
225×4=13AD2
AD2=225×413
AD=225×413
AD=15×213
AD=3013

∵CD=32AD
=32×303
=453

∴Area of ΔACD ≠Area of ΔABC=Area of rectangle ABCD
CD=4513=AB
AD=3013=BC

∴Area of 🗆ABCD=CD×AD
=4513×3013

Area of =45×301313
=135013
=1031113 cm2

∴Perimeter=(AB+BC)
=2(4513+3013
=2(45+3013)
=2×7513
=15013

Question 30

Sol :
(a)










From ΔBCD
By Pythagoras theorem
BD2=BC2+CD2
132=122+CD2
169=144+CD2
CD2=169144
CD2=25
CD=25=5

(i) sinϕ=CDBD
=513

(ii) tanθ=DEAE
=129=43


(b)











Tanθ=43
sinθ=DEAD
AD=12sinθ

cosθ=AEAD
AD=9cosθ

Question 31

Sol :
(i) 
LHS(sinA+cosA)2+(sinAcosA)2
sin2A+cos2A+2sinAcosA+sin2A+cos2A2sinAcosA

((a+b)2=a2+b2+2ab(ab)2=a2+b22ab)

2sin2A+2cos2A
2(sin2A+cos2A)
⇒2(1)=2
∴L.HS.=R.H.S

(ii) 
LHS
cot2A1sin2A+1
cos2Asin2A1sin2A+1
cos2A1+sin2Asin2A
(cos2A+sin2A)1sin2A
11sin2A
⇒0=RHS

(iii)
LHS
11+tan2A+11+cot2A
11+sin2Acos2A+11+cos2Asin2A
1cos2+sin2Acos2A+1sin2A+cos2Asin2A
cos2A(1)+sin2A(1)
cos2A+sin2A
⇒1=RHS

x2=a2cos2θ+b2sin2θ+2absinθcosθ...(1)
∵y=asinθ-bcosθ

Squaring on both sides
y2=(asinθbcosθ)2
y2=a2sin2θ+b2cos2θ2absinθcosθ...(2)

(1)+(2)
x2+y2=a2cos2θ+b2sin2θ+2absinθcosθ+a2sin2θ+b2cos2θ2absinθcosθ

=a2(cos2θ+sin2θ)+b2(sin2θ+cos2θ)
=a2(1)+b2(1)
=a2+b2
∴LHS=RHS

Question 32

Sol :
Given : 1sin2θ1cos2θ
cos2θsin2θ (sin2θ+cosθ=1)
cot2θ
cotθ

Question 33

Sol :
Given : sinθ+cosecθ=2
Squaring on both sides
(sinθ+cosecθ)2=22

sin2θ+cosec2θ+2sinθcosθ=4
sin2θ+cosec2θ+2sinθ1sinθ=4
sin2θ+cosec2θ=42
sin2θ+cosec2θ=2

Question 34

Sol :
Given : x=acosθ+bsinθ
y=asinθ-bcosθ

Squaring on both sides
x2=(acosθ+bsinθ)2

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