ML Aggarwal Solution Class 9 Chapter 18 Trigonometric Ratios and Standard Angles Exercise 18.2

 Exercise 18.2

Question 1

Sol :

(i) Given : cos18sin72

cos18sin(9018) sin(90θ)=cosθ

cos18cos18

=1


(ii) tan41cot49

tan41cot(9041) cot(90θ)=tanθ

tan41tan41

=1


(iii) Given : cosec1730sec7230 sec(90θ)=cosecθ

cosec1730sec(901730)

cosec1730cosec1730

=1


Question 2

Sol :

(i) 

cot40tan5012(cos35sin55)

cot40tan(9040)12[cos35sin(9035)]

cot40cot4012(cos35cos35)

112

212

12


(ii) (sin49cos41)2+(cos41sin49)2

[sin49cos(9049)]2+[cos41sin(9041)]2

(sin49sin49)2+(cos41cos41)2

12+12

⇒1+1=2


(iii) sin72cos18sec32cosec58

sin72cos(9012)sec32cosec(9032)

sin72sin72sec32sec32

⇒1-1=0


(iv) cos75sin15+sin12cos78cos18sin12

cos75sin(9075)+sin12cos(9012)cos18sin(9018)

cos75cos75+sin12sin12cos18cos18

⇒1+1-1

⇒1


(v) sin25sec65+cos25cosec65

sin25sec(9025)+cos25cosec(9025)

sin25cosec25+cos25sec25

sin251sin25+cos251cos25

sin225+cos225 (cos2θ+sin2θ=1)

⇒1


Question 3

(i) sin62cos28

sin62cos(9062)

sin62sin62

⇒0


(ii) cosec35sec55

cosec35sec(9035)

cosec35cosec35

⇒0


Question 4

Sol :

(i) cos226+cos64sin26+tan36cot54

cos226+cos(9026)sin26+tan36cot(9036)

cos226+sin26sin26+tan36tan36

cos226+sin226+1

⇒1+1=2


(ii) sec17cosec73+tan68cot22+cos244+cos246

sec17cosec(9017)+tan68cot(9068)+cos246+cos244

sec17sec17+tan68tan68+cos2(9044)+cos244

1+1+(sin244+cos244)

⇒2+1=3


Question 5

Sol :

(i) sin65cos25+cos32sin58sin28sec62+cosec230

sin65cos(9065)+cos32sin(9032)sin28sec(9028)+(2)2

sin65sin65+cos32cos32sin28cos28+4

1+1sin281sin28+4

⇒6-1=5


(ii) sec29cosec61+2cot8cot17cot45cot73cot823(sin238+sin252)

sec29cosec(9029)+2cot(9082)cot(9073)cot45cot73cot823(sin238+sin2(9038))

sec29cosec29+2cot82tan82cot73tan73cot453(sin238+cos238)

⇒1+2(1)-3(1)

⇒1+2-3=0


Question 6

Sol :

(i) tan81+cos72

⇒tan(90-9°)+cos(90-18°)

⇒cot 9°+sin 18°


(ii) cot49+cosec87

cot(9041)+cosec(905)

tan41+sec3


Question 7

Sol :

(i) sin228cos262=0

sin228cos262

sin228cos2(9028)

sin228sin228

⇒0

∴LHS=RHS

(ii) LHScos225+cos265

cos225+cos2(9025)

cos225+sin225

⇒1

=RHS

∴LHS=RHS


(iii) cosec267tan223

cosec2(9023)tan223

sec223tan223

⇒1

∴LHS=RHS


(iv) LHS sec222cot268

sec222cot2(9022)

sec222tan222

⇒1=RHS


Question 8

Sol :

(i) LHSsin63cos27+cos63sin27

sin63cos(9063)+cos63sin(9063)

sin63sin63+cos63cos63

sin263+cos263

⇒1

∴LHS=RHS


(ii) LHSsec31sin59+cos31cosec59

sec31sin(9031)+cos31cosec(9031)

sec31cos31+cos31sec31

2sec31cos31

21cos31.cos31

⇒2

∴LHS=RHS


Question 9

Sol :

(i) LHSsec70sin20cos20cossec70

sec70sin(9070)cos20cosec(9020)

sec70cos20cos28sec70

⇒0


(ii) LHSsin220+sin270tan245

sin220+sin2(9020)tan245

sin220+cos220(1)2

⇒1-1

⇒0

∴LHS=RHS


Question 10

Sol :

(i) LHScot54tan36+tan20cot702

cot54tan(9054)+tan20cot(9020)2

cot54cot54+tan20tan202

⇒1+1-2

⇒0

∴LHS=RHS


(ii) LHScos80sin10+cos59cosec31

cos80sin(9080)+cos59cosec(9031)

cos80cos80+cos89sec59

⇒1+1=2=RHS


Question 12

Sol :

(i) 2(tan35cot55)2+(cot55tan35)3(sec40cosec50)

2. [tan35cot(9035)]2+cot55tan(9055)3(sec40cosec(9040))

2(tan55tan35)2+cot55cot553sec40sec40

⇒2+1-3

⇒0


(ii) sin35cos55+cos35sin55cosec210tan280

sin35cos(9035)+cos35sin(9035)cosec2(9080)tan280

sin35sin35+cos35cos35sec280+tan280

sin235+cos235sec280tan280

11=1


(iii) sin234+sin256+2tan18tan72cot230

sin234+sin2(9034)+2tan18tan(9018)cot230

⇒1+2-3=0


Question 13

Sol :

(i) LHScosθsin(90θ)+sinθcos(90θ)

cosθcosθ+sinθsinθ

⇒1+1=2

∴LHS=RHS


(ii) L H Scosθsin(90θ)+sinθcos(90θ)

cosθcosθ+sinθsinθ

cos2θ+sin2θ

⇒1

∴LHS=RHS


(iii) LHS=tanθtan(90θ)+sin(90θ)cosθ

tanθcotθ+cosθcosθ

tanθ(1tanθ)+1

tan2θ+1 (sec2θtan2θ=1)

sec2θ

∴LHS=RHS


Question 14

Sol :

(i) LHS⇒cos(90A)sin(90A)tan(90A)

sinAcosAcotA

sinAcosAcosAsinA (sin2A+cos2A=1)

sin2A

1cos2A


(ii) LHSsin(90A)cosec(90A)+cos(90A)sec(90A)

sincosAsecA+sinAcosecA

cosA(1cosA)+sinA(1sinA)

cosAcosA+sinAsinA

cos2A+sin2A

⇒1

∴LHS=RHS


Question 15

Sol :

(i) cosθsin(90θ)+cos(90θ)sec(90θ)3tan230

cosθcosθ+sinAcosecθ3(13)2

1+sinA(1sinA)313

1+sin2A1

sin2A


(ii) cosec(90θ)sin(90θ)cot(90θ)cos(90θ)sec(90θ)tanθ+cotθtan(90θ)

secθcosθtanθsinθcosecθtanθ+cotθcotθ

1cosθcosθsinθ1sinθ+1

⇒1+1

⇒2


Question 16

Sol :

(i) cos63sec(90θ)=1

cos63cosecθ=1

cos63=1cosecθ.

cos63=sinθ

cos63=cos(90θ)

∴90-θ=63°

θ=90°-63°=27°


(ii) tan35cot(90θ)=1

cot(90θ)=1tan35

cot(90θ)=cot35

90θ=35

θ=9035

θ=55


(ii) LHS

4(sin430+cos460)3(cos245sin290)

4[(12)4+(12)4]3((12)212)

4(116+116)3(121)

4(1+116)3(12)

4216+32

12+32

12+32

1+32

42

⇒-2=RHS



(iii) LHS⇒cos60=12

RHScos230sin230=(32)2(12)2

=3414

=314

=24

=12

∴LHS=RHS

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