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=225−144
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
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) sec2A−tan2A
∵secA=1cosA
=11213
∵tanA=sinAcosA
=(513)(1213)
TanA=512
∴sec2A−tan2A=(1312)2−(512)2
=169144−25144
=169−25144
=144144
=1
=sec2A−tan2A=1
Question 2
Sol :
By Pythagoras theorem
∵ hypotenuse =BC
∴BC2=AB2+AC2
102=62+AC2
100=36+AC2
AC2=100−36
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+sinC⋅cosB
⇒45⋅45+35⋅35
⇒1625+925
⇒16+925
⇒2525
⇒1
(b) From Figure ΔADC
AD2+CD2=AC2
AD2+52=132
AD2+25=169
AD2=169−25
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) cosec2y−cot2y
⇒cosecy=ABAD
=2012
⇒coty=BDAD
=1612
coty=43
∴cosec2y−cot2y=(53)2−(43)2
=259−169
=25−169
=99
∴cosec2y−cot2y=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=169−25
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=144−16
=128
BD2=128
BD=√128
=√16×4×2
=4×2√2
BD=8√2
(i) Sin x=BCAC
sinx=45
(ii) cotx=ABBC
=34
(iii) cot2x−cosec2x
∵cotx=34
cosecx=ACBC
=54
∴cot2x−cosec2x=(34)2−(54)2
=916−254
∴cot2x−cosec2x=9−2516
=−1616
=−1
(iv) secy=CDBD
=128√2
secy =32√2
(v) tan2y−1cos2y
∵sec2y=1cos2y
=(32√2)2
=94×2
1cos2y=98
∴Tan2y=(BCBD)2
=(48√2)2
=1664×2
Tan2y=18
∴tan2y−1cos2y=18−98
=1−98
=−88
=-1
Question 4
Sol :
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=625−225
AB2=400
AB=√400=20
(i) 2sin y-cos y
∵siny=BDBC
=915
∴ Siny =35
cosy=CDBC
=1215
∴cosy=45
∴2siny−cosy=2⋅35−45
=65−45
=6−45=25
(ii) 2sin x-cos x
sinx=BCAC=1525=35
cosx=ABAC=2025=45
∴2sinx−cosx=2⋅35−45
=65−45
=6−45=25
(iii) 1-sin x+cos y
∵sinx=35
cosy=45
∴1−sinx+cosy
=1−35+45
=5−3+45
65
(iv) 2cosx-3siny+4tanx
∴sinx=35
cosx=45
siny=35
∴tanx=sinxcosx
=3545
tanx=34
∴2cosx-3siny+4tanx
⇒2⋅45−3⋅35+4⋅34
⇒85−95+3
⇒8−9+155
⇒145
(b) By Pythagoras theorem
(ii) AC2=BC2+AB2
52=32+AB2
25=9+AB2
AB2=25−9
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=135−125
=13−125=15
Question 6
(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
2500+QR2−100QR=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 :
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=43−34
=16−912=712
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=(2√5)2=45
tan2θ=(ABBC)2=(21)2=4
∴sin2θ+tan2θ=45+4
=4+205
=245
sin2θ+tan2θ=445
Given BD=15
Sin B=45
tan C=1
∵sinB=45×55
SinB=ADAB=2025
(i) AD=20
∵tanC⇒sinCcosC=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=20√2
(ii) tan2B−1cos2B=−1
∴LHS⇒tan2B−1cos2B
tan2B=(ADBD)2=(.2015)2=(43)2
=169
cos2B=(BDAB)2=(1525)2=(35)2=925
∴tan2B−1cos2B⇒169−1(925)
⇒169−259
⇒16−259
⇒−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=25−9
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
By Pythagoras theorem
bc2=ab2+ac2
102=62+ac2
100=36+ac2
ac2=100−36
ac2=64
ac=√64
ac=8
∴cosθ=acbc=810
tanθ=abac=68
∴cosθ+tanθ=810+68
=64+6080
=12480⇒3120
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=5−1
ac2=4
ac=√4
ac=2
∴cotθ−cosθ=acab−acbc
=21−2√5
=2√5−2√5
=2(√5−1)√5
Question 14
Sol :
Given : sinθ=pq
By Pythagoras theorem
q2=p2+ac2
ac2=a2−p2
ac=√q2−p2
cos θ=acbc=√−p2+q2q
∴cos θ +sin θ=√q2−p2q+pq
=p+√q2−p2q
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=169−25
AC2=144
AC=√144
AC=12
∴CB=5 ; BA=13 ; AC=12
sinA=513
cosA=1213
tanA=sinAcosA
=(513)(1213)
⇒512
∴5sinA−2cosAtanA=5(513)−2(1213)(512)
=2513−2413512
=25−2413512
=113×125=1265
Given : cosec A=√2
cosecA=1sinA=√2
∴sinA=1√2
By Pythagoras theorem
bc2=ab2+ac2
(√2)2=12+ac2
2=1+ac2
ac2=2−1
ac2=1
ac=√1
ac=1
∴ac=1 ; bc=√2 ; ab=1
∴sin A=1√2
cos A=1√2
tan A=sinAcosA
=(1√2)(1√2)=1
cot A1tanA
=11
cot A=1
∴2sin2A+3cot2Atan2A−cos2A
⇒2⋅(1√2)2+3(1)2(1)2−(1√2)2
⇒2⋅12+3⋅(1)1−12
⇒1+3(2−12)
⇒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+sinθcosθ−sinθ=1213+5131213−513
=12+51312−513
=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+cosA2cosA−sinA
=513+12132⋅1213−513
=(5+12)/13(24−5)/13
=1719
Question 21
Sol :
Given : tanθ=pq
By Pythagoras theorem
bc2=ab2+ac2
bc2=p2+q2
bc=√p2+q2
∴sin θ=abbc=p√p2+q2
cos θ=acbc=q√p2+q2
∴psinθ−qcosθpsinθ+qcosθ
=pp√p2+q2−q⋅q√p2+q2p⋅p√p2+q2+qq√p2+q2
=p2−q2√p2+q2p2+q2√p2+q2
p2−a2p2+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
∴5sinθ−3cosθ5sinθ+3cosθ
=5.35−3⋅455⋅35+3⋅45
=(15−12)/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+cosθ2cosθ−sinθ
=513+12132⋅1213−513
=5+121324−513
=1716
(ii)
Given : cosecθ=1312
⇒cosecθ=1sinθ=1312
∴sinθ=1213
By Pythagoras theorem
bc2=ab2+ac2
132=122+ac2
169=144+ac2
169−144=ac2
25=ac2
ac=√25
ac=5
∴sinθ=1213
cosθ=acbc=513
∴2sinθ−3cosθ4sinθ−9cosθ
=2⋅1213−3⋅5134⋅1213−9⋅513
=24−151348−4513
=93
=3
Question 24
Sol :
Given :
5sinθ=3
sinθ=35
By Pythagoras theorem
bc2=ab2+ac2
52=32+ac2
25=a+ac2
ac2=25−9
ac2=16
ac=√16
ac=4
cosθ=acbc=45
∴secθ=1cosθ=54
tanθ=sinθcosθ=3/54/5=34
∴secθ−tanθsecθ+tanθ
=54−3454+34
=5−345+34=28=14
Question 25
Sol :
Given :
sinθ=cosθ
Then 2tan2θ+sin2θ−1
⇒2(sinθcosθ)2+sin2θ−1
⇒2(sinθsinθ)2+sin2θ−1
⇒2(1)2+sin2θ−1
⇒2+sin2θ−1
⇒sin2θ+1
From given sinθ=cosθ
So, θ=45
∴sin45=cos45
1√2=1√2
∴2tan2θ+sin2θ−1
⇒sin2θ+1
⇒(1√2)2+1
⇒12+1=32
∴2tan2θ+sin2θ−1=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
⇒sinθ⋅cosθsinθ [∵cotθ=cosθsinθ]
⇒cosθ
∴L.H.S=R.H.S
(iii)
LHS⇒⇒sin2θcosθ+cos2θ
⇒sin2θ+cosθ⋅cosθcosθ
⇒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
cos⋅A=ACAB=45
cosB=BCAB=35
SinB=ACAB=45
∴sinA cosB+cosAsinB
⇒35⋅35+45⋅45
⇒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=15×2√13
AD=30√13
∵CD=32AD
=32×30√3
=45√3
∴Area of ΔACD ≠Area of ΔABC=Area of rectangle ABCD
∴CD=45√13=AB
AD=30√13=BC
∴Area of 🗆ABCD=CD×AD
=45√13×30√13
Area of =45×30√13⋅√13
=135013
=1031113 cm2
∴Perimeter=(AB+BC)
=2(45√13+30√13
=2(45+30√13)
=2×75√13
=150√13
Question 30
Sol :
(a)
From ΔBCD
By Pythagoras theorem
BD2=BC2+CD2
132=122+CD2
169=144+CD2
CD2=169−144
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+(sinA−cosA)2
⇒sin2A+cos2A+2sinAcosA+sin2A+cos2A−2sinAcosA
(∵(a+b)2=a2+b2+2ab(a−b)2=a2+b2−2ab)
⇒2sin2A+2cos2A
⇒2(sin2A+cos2A)
⇒2(1)=2
∴L.HS.=R.H.S
(ii)
LHS
⇒cot2A−1sin2A+1
⇒cos2Asin2A−1sin2A+1
⇒cos2A−1+sin2Asin2A
⇒(cos2A+sin2A)−1sin2A
⇒1−1sin2A
⇒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=(asinθ−bcosθ)2
y2=a2sin2θ+b2cos2θ−2absinθcosθ...(2)
(1)+(2)
x2+y2=a2cos2θ+b2sin2θ+2absinθcosθ+a2sin2θ+b2cos2θ−2absinθcosθ
=a2(cos2θ+sin2θ)+b2(sin2θ+cos2θ)
=a2(1)+b2(1)
=a2+b2
∴LHS=RHS
Question 32
Sol :
Given : √1−sin2θ1−cos2θ
⇒√cos2θsin2θ (∵sin2θ+cosθ=1)
⇒√cot2θ
⇒cotθ
Question 33
Sol :
Given : sinθ+cosecθ=2
Squaring on both sides
(sinθ+cosecθ)2=22
sin2θ+cosec2θ+2⋅sinθ⋅cosθ=4
sin2θ+cosec2θ+2sinθ1sinθ=4
sin2θ+cosec2θ=4−2
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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