ML Aggarwal Solution Class 9 Chapter 18 Trigonometric Ratios and Standard Angles Exercise 18.2
Exercise 18.2
Question 1
Sol :
(i) Given : cos18∘sin72∘
⇒cos18∘sin(90∘−18∘) ∵sin(90−θ)=cosθ
⇒cos18∘cos18∘
=1
(ii) tan41∘cot49∘
⇒tan41∘cot(90∘−41∘) ∵cot(90∘−θ)=tanθ
⇒tan41∘tan41∘
=1
(iii) Given : cosec17∘30′sec72∘30′ sec(90−θ)=cosecθ
⇒cosec17∘30′sec(90∘−17∘30′)
⇒cosec17∘30′cosec17∘30′
=1
Question 2
Sol :
(i)
cot40∘tan50∘−12(cos35∘sin55∘)
⇒cot40∘tan(90−40∘)−12[cos35∘sin(90−35∘)]
⇒cot40∘cot40∘−12(cos35∘cos35∘)
⇒1−12
⇒2−12
⇒12
(ii) (sin49∘cos41∘)2+(cos41∘sin49∘)2
⇒[sin49∘cos(90−49∘)]2+[cos41∘sin(90−41)]2
⇒(sin49∘sin49∘)2+(cos41∘cos41)2
⇒12+12
⇒1+1=2
(iii) sin72∘cos18∘−sec32∘cosec58∘
⇒sin72∘cos(90−12∘)−sec32∘cosec(90−32∘)
⇒sin72sin72∘−sec32∘sec32∘
⇒1-1=0
(iv) cos75∘sin15∘+sin12∘cos78∘−cos18∘sin12∘
⇒cos75∘sin(90−75∘)+sin12∘cos(90−12∘)−cos18∘sin(90−18∘)
⇒cos75∘cos75∘+sin12∘sin12∘−cos18∘cos18∘
⇒1+1-1
⇒1
(v) sin25∘sec65∘+cos25∘cosec65∘
⇒sin25∘sec(90−25)+cos25∘cosec(90−25∘)
⇒sin25∘cosec25∘+cos25∘sec25∘
⇒sin25∘1sin25∘+cos25∘1cos25∘
⇒sin225∘+cos225∘ (∵cos2θ+sin2θ=1)
⇒1
Question 3
(i) sin62∘−cos28∘
⇒sin62∘−cos(90−62∘)
⇒sin62∘−sin62∘
⇒0
(ii) cosec35∘−sec55∘
⇒cosec35∘−sec(90−35∘)
⇒cosec35∘−cosec35∘
⇒0
Question 4
Sol :
(i) cos226∘+cos64∘sin26∘+tan36∘cot54∘
⇒cos226∘+cos(90−26∘)sin26∘+tan36∘cot(90−36∘)
⇒cos226∘+sin26∘sin26∘+tan36∘tan36∘
⇒cos226∘+sin226+1
⇒1+1=2
(ii) sec17∘cosec73∘+tan68∘cot22∘+cos244∘+cos246∘
⇒sec17∘cosec(90−17)+tan68∘cot(90−68∘)+cos246∘+cos244∘
⇒sec17∘sec17∘+tan68∘tan68∘+cos2(90∘−44∘)+cos244∘
⇒1+1+(sin244∘+cos244)
⇒2+1=3
Question 5
Sol :
(i) sin65∘cos25∘+cos32∘sin58∘−sin28∘sec62∘+cosec230∘
⇒sin65∘cos(90−65∘)+cos32∘sin(90−32∘)−sin28sec(90−28∘)+(2)2
⇒sin65∘sin65∘+cos32∘cos32∘−sin28⋅cos28+4
⇒1+1−sin281sin28+4
⇒6-1=5
(ii) sec29∘cosec61∘+2cot8∘⋅cot17∘⋅cot45∘⋅cot73∘⋅cot82∘−3(sin238∘+sin252∘)
⇒sec29∘cosec(90−29∘)+2cot(90−82∘)⋅cot(90−73∘)cot45∘cot73∘cot82∘−3(sin238∘+sin2(90−38∘))
⇒sec29∘cosec29∘+2cot82∘tan82∘⋅cot73∘⋅tan73∘cot45∘−3(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(90−41∘)+cosec(90−5∘)
⇒tan41∘+sec3∘
Question 7
Sol :
(i) sin228∘−cos262∘=0
⇒sin228−cos262∘
⇒sin228∘−cos2(90−28∘)
⇒sin228∘−sin228∘
⇒0
∴LHS=RHS
(ii) LHScos225∘+cos265∘
cos225∘+cos2(90−25∘)
cos225∘+sin225
⇒1
=RHS
∴LHS=RHS
(iii) ⇒cosec267∘−tan223∘
⇒cosec2(90−23∘)−tan223∘
⇒sec223∘−tan223∘
⇒1
∴LHS=RHS
(iv) LHS sec222∘−cot268∘
⇒sec222∘−cot2(90−22∘)
⇒sec222∘−tan222∘
⇒1=RHS
Question 8
Sol :
(i) LHS⇒sin63∘⋅cos27∘+cos63∘sin27∘
⇒sin63∘cos(90−63∘)+cos63∘sin(90−63∘)
⇒sin63∘sin63∘+cos63∘⋅cos63∘
⇒sin263∘+cos263∘
⇒1
∴LHS=RHS
(ii) LHSsec31∘sin59∘+cos31∘cosec59∘
⇒sec31∘sin(90−31∘)+cos31∘cosec(90−31∘)
⇒sec31∘cos31∘+cos31∘sec31∘
⇒2sec31∘cos31∘
⇒21cos31∘.cos31∘
⇒2
∴LHS=RHS
Question 9
Sol :
(i) LHS⇒sec70∘sin20∘−cos20∘⋅cossec70∘
⇒sec70∘sin(90−70∘)−cos20∘cosec(90−20∘)
⇒sec70∘⋅cos20∘−cos28∘sec70∘
⇒0
(ii) LHS⇒sin220∘+sin270∘−tan245
⇒sin220∘+sin2(90−20∘)−tan245∘
⇒sin220+cos220∘−(1)2
⇒1-1
⇒0
∴LHS=RHS
Question 10
Sol :
(i) LHS⇒cot54∘tan36∘+tan20∘cot70∘−2
⇒cot54∘tan(90−54)+tan20∘cot(90−20)−2
⇒cot54∘cot54∘+tan20∘tan20∘−2
⇒1+1-2
⇒0
∴LHS=RHS
(ii) LHS⇒cos80∘sin10∘+cos59∘cosec31∘
⇒cos80∘sin(90−80∘)+cos59∘cosec(90−31)∘
⇒cos80∘cos80∘+cos89∘sec59∘
⇒1+1=2=RHS
Question 12
Sol :
(i) 2(tan35∘cot55∘)2+(cot55∘tan35∘)−3(sec40∘cosec50∘)
⇒ 2. [tan35∘cot(90−35∘)]2+cot55∘tan(90−55∘)−3(sec40∘cosec(90−40∘))
⇒2(tan55∘tan35∘)2+cot55cot55−3sec40sec40
⇒2+1-3
⇒0
(ii) sin35∘cos55∘+cos35∘sin55∘cosec210∘−tan280
⇒sin35∘⋅cos(90−35∘)+cos35∘sin(90−35∘)cosec2(90−80∘)−tan280∘
⇒sin35∘⋅sin35∘+cos35∘⋅cos35∘sec280∘+tan280∘
sin235∘+cos235∘sec280−tan280
⇒11=1
(iii) sin234∘+sin256∘+2tan18∘⋅tan72∘−cot230∘
⇒sin234∘+sin2(90−34)+2tan18⋅tan(90−18∘)−cot230
⇒1+2-3=0
Question 13
Sol :
(i) LHS⇒cosθsin(90−θ)+sinθcos(90−θ)
⇒cosθcosθ+sinθsinθ
⇒1+1=2
∴LHS=RHS
(ii) L H S⇒cosθ⋅sin(90−θ)+sinθ⋅cos(90−θ)
⇒cosθ⋅cosθ+sinθ⋅sinθ
⇒cos2θ+sin2θ
⇒1
∴LHS=RHS
(iii) LHS=tanθtan(90−θ)+sin(90−θ)cosθ
⇒tanθcotθ+cosθcosθ
⇒tanθ(1tanθ)+1
⇒tan2θ+1 (∵sec2θ−tan2θ=1)
⇒sec2θ
∴LHS=RHS
Question 14
Sol :
(i) LHS⇒cos(90∘−A)⋅sin(90∘−A)tan(90∘−A)
⇒sinA⋅cosAcotA
⇒sinA⋅cosAcosAsinA (∵sin2A+cos2A=1)
⇒sin2A
⇒1−cos2A
(ii) LHS⇒sin(90∘−A)cosec(90∘−A)+cos(90∘−A)sec(90∘−A)
⇒sincosAsecA+sinAcosecA
⇒cosA(1cosA)+sinA(1sinA)
⇒cosA⋅cosA+sinA⋅sinA
⇒cos2A+sin2A
⇒1
∴LHS=RHS
Question 15
Sol :
(i) cosθsin(90−θ)+cos(90∘−θ)sec(90∘−θ)−3tan230∘
⇒cosθcosθ+sinAcosecθ−3(1√3)2
⇒1+sinA(1sinA)−313
⇒1+sin2A−1
⇒sin2A
⇒secθ⋅cosθ⋅tanθsinθ⋅cosecθ⋅tanθ+cotθcotθ
1cosθ⋅cosθsinθ1sinθ+1
⇒1+1
⇒2
Question 16
Sol :
(i) cos63∘⋅sec(90∘−θ)=1
⇒cos63∘cosecθ=1
⇒cos63∘=1cosecθ.
⇒cos63∘=sinθ
⇒cos63∘=cos(90−θ)
∴90-θ=63°
θ=90°-63°=27°
(ii) tan35∘⋅cot(90−θ)=1
cot(90∘−θ)=1tan35∘
cot(90∘−θ)=cot35∘
90∘−θ=35∘
θ=90∘−35∘
θ=55∘
(ii) LHS
⇒4(sin430∘+cos460∘)−3(cos245∘−sin290∘)
⇒4⋅[(12)4+(12)4]−3((1√2)2−12)
⇒4(116+116)−3(12−1)
⇒4(1+116)−3(−12)
⇒4216+32
⇒12+32
⇒12+32
⇒1+32
⇒−42
⇒-2=RHS
(iii) LHS⇒cos60∘=12
RHS⇒cos230∘−sin230∘=(√32)2−(12)2
=34−14
=3−14
=24
=12
∴LHS=RHS
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