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Page No 339:

Question 1:

Without using trigonometric tables, evaluate:
(i) sin 16°cos 74°
(ii) sec 11°cosec 79°
(iii) tan 27°cot 63°
(iv) cos 35°sin 55°
(v) cosec 42°sec 48°
(vi) cot 38°tan 52°

Answer:

(i)sin16cos74  =sin(9074)cos74              =cos74cos74            [sin 90-θ = cos θ]=1(ii) sec11cosec79      =sec(9079)cosec79        =cosec79cosec79         [sec 90-θ = cosec θ]=1      (iii)tan27cot63 =tan(9063)cot63        =cot63cot63     [tan 90-θ = cot θ]   =1(iv)cos35sin55    =cos(9055)sin55           =sin55sin55      [sin 90-θ = cosθ]    =1( v)cosec42sec48    =cosec(9048)sec48        =sec48sec48        [sec 90-θ = cosec θ]    =1(vi)cot38tan52    =cot(9052)tan52         =tan52tan52           [tan 90-θ = cot θ] =1



Page No 340:

Question 2:

Prove that:

(i) sinθ cos(90°-θ)+sin(90°-θ) cosθ=1
(ii) sinθcos(90°-θ)+cosθsin(90°-θ)=2
(iii) sinθ cos(90°-θ)cosθsin(90°-θ)+cosθ sin(90°-θ)sinθcos(90°-θ)
(iv) cos(90°-θ)sec(90°-θ)tanθcosec(90°-θ)sin(90°-θ)cot(90°-θ)+tan(90°-θ)cotθ=2
(v) cos(90°-θ)1+sin(90°-θ)+1+sin(90°-θ)cos(90°-θ)=2cosecθ
(vi) sec90°-θ cosecθ-tan90°-θ cotθ+cos225°+cos265°3tan27° tan63°=23           CBSE 2010
(vii) cotθ tan90°-θ-sec90°-θcosecθ+3tan12° tan60° tan78°=2       CBSE 2010

Answer:

(i) LHS=sinθcos(900θ)+sin(900θ)cosθ                        =sinθsinθ+cosθcosθ                        =sin2θ+cos2θ                        =1                         = RHS            Hence proved.(ii) LHS=sinθcos(900θ)+cosθsin(900θ)              =sinθsinθ+cosθcosθ              =1+1              =2             =RHS        Hence proved.  (iii) LHS=sinθcos(900θ)cosθsin(900θ)+cosθsin(900θ)sinθcos(900θ)                =sinθsinθcosθcosθ+cosθcosθsinθsinθ                =sin2θ+cos2θ                =1            =RHS            Hence proved.             (iv) LHS=cos(900θ)sec(900θ)tanθcosec(900θ)sin(900θ)cot(900θ)+tan(900θ)cotθ                =sinθcosecθtanθsecθcosθtanθ+cotθcotθ                         =1+1                =2   =RHS            Hence proved.(v) LHS=cos(900θ)1+sin(900θ)+1+sin(900θ)cos(900θ)              =sinθ1+cosθ+1+cosθsinθ              =sin2θ+(1+cosθ)2(1+cosθ)sinθ              =sin2θ+1+cos2θ+2cosθ(1+cosθ)sinθ              =1+1+2cosθ(1+cosθ)sinθ              =2+2cosθ(1+cosθ)sinθ              =2(1+cosθ)(1+cosθ)sinθ               =21sinθ               =2cosecθ            = RHS            Hence proved.

vi LHS=sec90°-θ cosecθ-tan90°-θ cotθ+cos225°+cos265°3tan27° tan63°=cosecθ cosecθ-cotθ cotθ+sin290°-25°+cos265°3tan27° cot90°-63°=cosec2θ-cot2θ+sin265°+cos265°3tan27° cot27°=1+13×tan27°×1tan27°=23=RHS

vii LHS=cotθ tan90°-θ-sec90°-θcosecθ+3tan12° tan60° tan78°=cotθ cotθ-cosecθ cosecθ+3tan12°×3×cot90°-78°=cot2θ-cosec2θ+3tan12° cot12°=-1+3×tan12°×1tan12°=-1+3=2=RHS

Page No 340:

Question 3:

Without using trigonometric tables, evaluate:
(i) sin50°cos40°+cosec40°sec50°-4 cos52° cosec38°
(ii) cos75°sin15°+sin12°cos78°-cos18° cosec72°
(iii) 2cosec67°sec23°-tan70°cot20°-cos0°+tan38° tan52°
(iv) tan76°cot14°+sec58°cosec32°-sin35° sec55°-8sin230°

Answer:

(i)sin500cos400+cosec400sec5004cos520cosec380   =sin(900400)cos400+cosec(900500)sec5004cos(900380)cosec380          =cos400cos400+sec500sec5004sin3801sin380             =1+14          =2    (ii)cos750sin150+sin120cos780cos180cosec720        =cos(900150)sin150+sin(900780)cos780cos(900720)cosec720       =sin150sin150+cos780cos780sin720.1sin720       =1+11  =1(iii)2cosec670sec230tan700cot200cos00+tan380tan520      =2cosec(900230)sec230tan(900200)cot2001+tan(900520)tan520      =2sec230sec230cot200cot2001+cot520.1cot520      =211+1      =1(iv)tan760cot140+sec580cosec320sin350sec5508sin2300     =tan(900140)cot140+sec(900320)cosec320sin(900550)sec5508×(12)2     =cot140cot140+cosec320cosec320cos550.1cos5508×14     =1+112=1

Page No 340:

Question 4:

Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 80° − sec 10° = 0
(iv) cosec272° − tan218° = 1
(v) cos275° + cos215° = 1
(vi) tan266° − cot224° = 0
(vii) sin248° + sin242° = 1
(viii) cos257° − sin233° = 0
(ix) (sin 65° + cos 25°)(sin 65° − cos 25°) = 0

Answer:

(i) LHS=cos810sin90              =cos(90090)sin90              =sin90sin90              =0              =RHS(ii) LHS=tan710cot190               =tan(900190)cot190               =cot190cot190              =0=RHS(iii) LHS=cosec800sec100                =cosec(900100)sec100                =sec100sec100                =0               =RHS(iv) LHS=cosec2720tan2180                =cosec2(900180)tan2180                =sec2180tan2180                =1     =RHS(v) LHS=cos2750+cos2150               =cos2(900150)+cos2150               =sin2150+cos2150               =1.  =RHS(vi) LHS=tan2660cot2240                =tan2(900240)cot2240                =cot2240cot2240                =0=RHS(vii) LHS=sin2480+sin2420                 =sin2(900420)+sin2420                 =cos2420+sin2420                 =1                 =RHS(viii) LHS=cos2570sin2330                  =cos2(900330)sin2330                  =sin2330sin2330                 =0  =RHS(ix) LHS=(sin650+cos250)(sin650cos250)               =sin2650cos2250               =sin2(900250)cos2250               =cos2250cos250               =0  =RHS

Page No 340:

Question 5:

Without using trigonometric tables, prove that:

(i) sin53° cos37° + cos53° sin37° = 1
(ii) cos54° cos36° − sin54° sin36° = 0
(iii) sec70° sin20° + cos20° cosec70° = 2
(iv) sin35° sin55° − cos35° cos55° = 0
(v) (sin72° + cos18°)(sin72° − cos18°) = 0
(vi) tan48° tan23° tan42° tan67° = 1

Answer:

(i) LHS=sin530cos370+cos530sin370           =sin (900370)cos370+cos(900370)sin370             =cos370cos370+sin370sin370             =cos2370+sin2370             =1=RHS(ii) LHS=cos540cos360sin540sin360              =cos(900360)cos360sin(900360)sin360           =sin360cos360cos360sin360              =0=RHS(iii) LHS=sec700sin200+cos200cosec700               =sec(900200)sin200+cos200cosec(900200)               =cosec200.1cosec200+1sec200.sec200             =1+1           =2=RHS

iv LHS=sin35° sin55°-cos35° cos55°=sin35° cos90°-55°-cos35° sin90-55°=sin35° cos35°-cos35° sin35°=0=RHS

v LHS=sin72°+cos18°sin72°-cos18°=sin72°+cos18°cos90°-72°-cos18°=sin72°+cos18°cos18°-cos18°=sin72°+cos18°0=0=RHS

vi LHS=tan48° tan23° tan42° tan67°=cot90°-48° cot90°-23° tan42° tan67°=cot42° cot67° tan42° tan67°=1tan42°×1tan67°×tan42°×tan67°=1=RHS

Page No 340:

Question 6:

Prove that:

(i) tan5° tan25° tan30° tan65° tan85° = 13
(ii) cot12° cot38° cot52° cot60° cot78° = 13
(iii) cos15° cos35° cosec55° cos60° cosec75° = 12
(iv) cos1° cos2° cos3° ... cos180° = 0
(v) sin49°cos41°2+cos41°sin49°2=2

Answer:

 (i) LHS=tan50tan250tan300tan650tan850               =tan(900850)tan(900650)×13×1cot6501cot850               =cot850cot650131cot6501cot850               =13=RHS

ii LHS=cot12° cot38° cot52° cot60° cot78°=tan90°-12°×tan90°-38°×cot52°×13×cot78°=13×tan78°×tan52°×cot52°×cot78°=13×tan78°×tan52°×1tan52°×1tan78°=13=RHS

(iii) LHS=cos150cos350cosec550cos600cosec750               =cos(900750)cos(900550)1sin550×12×1sin750               =sin750sin5501sin550×12×1sin750              =12=RHS

iv LHS=cos1° cos2° cos3° ... cos180°=cos1°×cos2°×cos3°×...×cos90°×...cos180°=cos1°×cos2°×cos3°×...×0×...cos180°=0=RHS

v LHS=sin49°cos41°2+cos41°sin49°2=cos90°-49°cos41°2+cos41°cos90°-49°2=cos41°cos41°2+cos41°cos41°2=12+12=1+1=2=RHS

Disclaimer: The RHS of (v) given in textbook is incorrect. There should be 2 instead 1. The same has been corrected in the solution here.



Page No 341:

Question 7:

Prove that
(i) sin (70° + θ) − cos (20° − θ) = 0
(ii) tan (55° − θ) − cot (35° + θ) = 0
(iii) cosec (67° + θ) − sec (23° − θ) = 0
(iv) cosec (65 °+ θ)  sec  (25° −  θ) − tan (55° − θ) + cot (35° + θ) = 0
(v) sin (50° + θ ) − cos (40° − θ) + tan 1° tan 10° tan 80° tan 89° = 1.

Answer:

(i) L.H.S=sin(700+θ)cos(200θ)             =sin{900(200θ)}cos(200θ)             =cos(200θ)cos(200θ)             =0=R.H.S.(ii) L.H.S=tan(550θ)cot(350+θ)              =tan{900(350+θ)}cot(350+θ)              =cot(350+θ)cot(350+θ)              =0=R.H.S.(iii) L.H.S=cosec(670+θ)sec(230θ)               =cosec{900(230θ)}sec(230θ)              =sec(230θ)sec(230θ)              =0=R.H.S.(iv) L.H.S=cosec(650+θ)sec(250θ)tan(550θ)+cot(350+θ)              =cosec{900(250θ)}sec(250θ)tan(550θ)+cot{900(550θ)}              =sec(250θ)sec(250θ)tan(550θ)+tan(550θ)              =0= R.H.S(v) L.H.S=sin(500+θ)cos(400θ)+tan10tan100tan800tan890              =sin{900(400θ)}cos(400θ)+{tan10tan(90010)}{tan100tan(90010)}          =cos(400θ)cos(400θ)+(tan10cot10)(tan100cot100)              =(1cot10×cot10)(tan100×1tan100)             =1×1             =1=R.H.S

Page No 341:

Question 8:

Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

(i) sin67° + cos75°
(ii) cot65° + tan49°
(iii) sec78° + cosec56°
(iv) cosec54° + sin72°

Answer:

i sin67°+cos75°=cos90°-67°+sin90°-75°=cos23°+sin15°

ii cot65°+tan49°=tan90°-65°+cot90°-49°=tan25°+cot41°

iii sec78°+cosec56°=sec90°-12°+cosec90°-34°=cosec12°+sec34°

iv cosec54°+sin72°=sec90°-54°+cos90°-72°=sec36°+cos18°

Page No 341:

Question 9:

If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.

Answer:

 sin3A=cos(A260)       ⇒cos(9003A)=cos(A260)       [sinθ=cos(900θ)]      9003A=A260      1160=4A     A=11604=290

Page No 341:

Question 10:

If tan 2 A = cot (A − 12°), where 2 A is an acute angle, find the value of A.

Answer:

  tan2A=cot(A120)         =>cot(9002A)=cot(A120)   [tanθ=cot(900θ)]         =>(9002A)=(A120)         =>1020=3A        =>A=10203=340

Page No 341:

Question 11:

If sec 4 A = cosec (A − 15°), where 4 A is an acute angle, find the value of A.

Answer:

   sec4A=cosec(A150)          => cosec(9004A)=cosec(A150)   [secθ=cosec(900θ)]          =>9004A=A150         =>1050=5A         =>A=10505=210

Page No 341:

Question 12:

If A, B, C are the angles of a ΔABC, prove that tan (B+C2) = cot A2.

Answer:

  We know that the sum of the angles of a triangle is 1800. A+B+C=1800                   =>B+C=1800A            =>B+C2=900A2                 [Dividing both sides by 2]           =>tan(B+C2)=tan(900A2)          =>tan(B+C2)=cotA2       [tan(900θ)=cotθ]       Hence proved.    

Page No 341:

Question 13:

sin15° cos75°+cos15° sin75°tan5° tan30° tan 35° tan 55° tan85°

Answer:

sin150cos750+cos150sin750tan50tan300tan350tan550tan850        =sin150cos(900150)+cos150sin(900150)tan300{tan50tan350tan550tan850}      =sin150sin150+cos150cos15013{tan(90050)tan850.tan350tan(900350)}     =sin2150+cos215013(cot850tan850)(tan350.cot350)   =3(1tan850tan850)(1cot350.cot350)  =31×1    =3

Page No 341:

Question 14:

3cos55°7sin35°-4(cos70° cosec20°)7(tan5° tan25° tan45° tan65° tan85°)

Answer:

3cos5507sin3504(cos700cosec200)7(tan50tan250tan450tan650tan850)   =3cos(900350)7sin3504(cos7001sin200)7×1{tan50tan(90050).tan250tan(900650)}     =3sin357sin3504(cos7001sin(900700))7(tan50cot50)(tan250cot250)     =374(cos7001cos700)7(tan501tan50)(tan2501tan250)    =3747    =347=17

Page No 341:

Question 15:

Prove that:

23cosec258°-23cot58° tan32°-53tan13° tan37° tan45° tan53° tan77°=-1

Answer:

23cosec258023cot580tan32053tan130tan370tan450tan530tan770      =23(cosec2580cot580tan320)53tan130tan(900130)tan370tan(900370)(tan450)     =23{cosec2580cot580tan(900580)}53tan130cot130tan370cot370(1)    =23(cosec2580cot580cot580)53tan1301tan130tan3701tan370     =23(cosec2580cot2580)53    =2353     =1

Hence Proved

Page No 341:

Question 16:

cos70°sin20°+cos55° cosec35°tan5° tan25° tan45° tan65° tan85°

Answer:

cos700sin200+cos550cosec350tan50tan250tan450tan650tan850        =cos(900200)sin200+cos550cosec(900550)tan450tan250tan(900250)tan(90050)tan50      =sin200sin200+cos550sec5501×tan250cot250cot50tan50     =1+cos5501cos550tan2501tan250cot501cot50     =1+11=1+1=2

Page No 341:

Question 17:

sin70°cos20°+cosec36°sec54°-2cos43° cosec47°tan10° tan40° tan50° tan80°

Answer:

 sin700cos200+cosec360sec5402cos430cosec470tan100tan400tan500tan800      =sin(900200)cos200+cosec(900540)sec5402cos430cosec(900430)tan100tan(900100)tan400tan(900400)   =cos200cos200+sec540sec5402cos430sec430tan100cot100tan400cot400   =1+12cos4301cos430tan1001tan100tan4001tan400  =22   =0

Page No 341:

Question 18:

-tanθ·cot(90°-θ)+secθ cosec(90°-θ)+sin235°+sin255°tan10° tan20° tan30° tan70° tan80°

Answer:

tanθcot(900θ)+secθcosec(900θ)+sin2350+sin2550tan100tan200tan300tan700tan800         =tanθtanθ+secθsecθ+sin2350+sin2(900350){tan100tan(900100)tan200tan(900200)}tan300          =tan2θ+sec2θ+(sin2350+cos2350)(tan100cot100)(tan200cot200)13        =sec2θtan2θ+1(tan1001tan100)(tan2001tan200)13        =(1+1)31       =23



Page No 342:

Question 19:

tan 7° tan 23° tan 60° tan 67° tan 83° + cot54°tan36° + sin 20° sec 70° − 2

Answer:

  tan70tan230tan600tan670tan830+cot540tan360+sin200sec7002         =(tan600)(tan70tan830)(tan230tan670)+cot(900360)tan360+sin200sec(900200)2          =3{tan70tan(90070)}{tan230tan(900230)}+tan360tan360+sin200cosec2002         =3(tan70cot70)(tan230cot230)+1+sin2001sin2002        =3×(1×1)+1+1-2        =3×1        =3

Page No 342:

Question 20:

(sin225° + sin265°) + 3 (tan 5° tan 15° tan 30° tan 75° tan 85°)

Answer:

(sin2250+sin2650)+3(tan50tan150tan300tan750tan850) =2250+sin2(900250)+3(tan300)(tan50tan850)(tan150tan750)       =(sin2250+cos2250)+3×13{tan50tan(90050)}{tan150tan(900150)}       =1+1(tan50cot50)(tan150cot150)        =1+tan501tan50tan1501tan150       =1+1       =2

Page No 342:

Question 21:

sin240°+sin250°cos220°+cos270°+tan10° tan20° tan60° tan70° tan80°

Answer:

sin2400+sin2500cos2200+cos2700+tan100tan200tan600tan700tan800         =sin2400+sin2(900400)cos2200+cos2(900200)+(tan600)tan200tan(900200)tan100tan(900100)        =sin2400+cos2400cos2200+sin2200+3(tan200cot200tan100cot100)       =1+3(tan2001tan200tan1001tan100)       =(1+3)      

Page No 342:

Question 22:

sec254°-cot236°cosec257°-tan233°+2sin238° sec252°-sin245°

Answer:

sec2540cot2360cosec2570tan2330+2sin2380sec2520sin2450         =sec2540cot2(900540)cosec2570tan2(900570)+2sin2380sec2(900380)(12)2      =sec2540tan2540cosec2570cot2570+2sin2380cosec238012       =1+2sin23801sin238012      =1+212         =312 =52

Page No 342:

Question 23:

sec2(90°-θ)-cot2θ2(sin225°+sin265°)+2cos260° tan228° tan262°3(sec243°-cot247°)

Answer:

sec2(900θ)cot2θ2(sin2250+sin2650)+2cos2600tan2280tan2623(sec2430cot2470)   =cosec2θcot2θ2{sin2250+sin2(900250)}+2(12)2tan2280tan2(900280)3{sec2430cot2(900430)}  =12(sin2250+cos2250)+2×14tan2280cot22803(sec2430tan2430)  =12+12×tan22801tan22803 =12+16  =3+16  =46=23

Page No 342:

Question 24:

sec2θ-cot2(90°-θ)cosec267°-tan223°+(sin240°+sin250°)

Answer:

sec2θcot2(900θ)cosec2670tan2230+(sin2400+sin2500)    =sec2θtan2θcosec2670tan2(900670)+sin2400+sin2(900400)   =1cosec2670cot2670+(sin2400+cos2400)   =1+1  =2

Page No 342:

Question 25:

sec254°-cot236°cosec257°-tan233°+2sin238° sec252°-sin245°

Answer:

sec2540cot2360cosec2570tan2330+2sin2380sec2520sin2450  =sec2540cot2(900540)cosec2570tan2(900570)+2sin2380sec2(900380)sin2450 =sec2540tan2540cosec2570cot2570+2sin2380cosec2380(12)2 =1+2sin23801sin238012 =1+212 =312 =52

Page No 342:

Question 26:

3tan25° tan40° tan50° tan65°-12tan260°4(cos229°+cos261°)

Answer:

3tan250tan400tan500tan65012tan26004(cos2290+cos2610)   =3tan250tan(900250)tan400tan(900400)12(3)24{cos229+cos2(900290)}  =3tan250cot250tan400cot400324(cos2290+sin2290)  =3tan2501tan250tan4001tan400324 =3324 =6324 =638=38

Page No 342:

Question 27:

23cosec258°-23cot58° tan32°                   -53tan13° tan37° tan45° tan53° tan77°

Answer:

23cosec258023cot580tan32053tan130tan370tan450tan530tan770      =23(cosec2580cot580tan320)53tan130tan(900130)tan370tan(900370)(tan450)     =23{cosec2580cot580tan(900580)}53tan130cot130tan370cot370(1)    =23(cosec2580cot580cot580)53tan1301tan130tan3701tan370     =23(cosec2580cot2580)53    =2353     =1



Page No 343:

Question 1:

tan30°cot60°=?
(a) 12
(b) 13
(c) 3
(d) 1

Answer:

(d) 1

tan 30°cot 60°   = tan 30°tan 60°  tan θ = 1cotθ = 13×3 =1

Page No 343:

Question 2:

tan35°cot55°+cot78°tan12°=?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

(c) 2We have: tan350cot550+cot780tan12o=tan350cot(900350)+cot(900120)tan120=tan350tan350+tan120tan120         [cot(900θ)=tanθ]=1+1=2

Page No 343:

Question 3:

tan25°cosec65°2+cot25°sec25°2=?
(a) 1
(b) 2
(c) 34
(d) 8

Answer:

(a) 1We have: tan250cosec6502+cot250sec65o2=[tan(900650)cosec650]2+[cot(900650)sec650]2     [tan(900θ)=cotθ and cot(900θ)=tanθ]=(cos650sin650×sin650)2+(sin650cos650×cos650)2=cos2650+sin2650=1

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Question 4:

cos 1° cos 2° cos 3° ... cos 180° = ?
(a) −1
(b) 1
(c) 0
(d) 12

Answer:

(c) 0cos10cos20cos30...cos1800=cos10cos20cos30...cos 900...cos(180)0=0     [cos 90° =0]

Page No 343:

Question 5:

tan 10° tan 15° tan 75° tan 80° = ?
(a) 3
(b) 13
(c) −1
(d) 1

Answer:

(d) 1We have:   tan100tan150tan750tan800=tan100×tan150×tan(900150)×tan(900100)=tan100×tan150×cot150×cot100                         [tan(900θ)=cotθ]=1

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Question 6:

tan 1° tan 2° tan 3° ... tan 89° = ?
(a) 1
(b) 0
(c) Cannot be determined
(d) None of these

Answer:

(a) 1  We have:tan10tan20tan30...tan890=tan10tan20...tan450...tan880tan890=tan10tan20...tan450...tan880tan890=tan10tan20...tan450...tan(90020)tan(90010)=tan10tan20...cot20cot10      [tan(900θ)=cotθ and tan450=1]=1cot10×1cot20...1...cot20×cot10=1

Page No 343:

Question 7:

tan 5° tan 25° tan 30° tan 65° tan 85° = ?
(a) 3
(b) 13
(c) 1
(d) none of these

Answer:

The correct option is (b).We have:tan50tan250tan300.tan650tan850= tan50tan250tan300tan(900250) tan(90050)=tan50tan250×13×cot250cot50      [tan(900θ)=cotθ and tan300=13]=13

Page No 343:

Question 8:

cot 15° cot 16° cot 17° ... cot 73° cot 74° cot 75° = ?
(a) 12
(b) 0
(c) 1
(d) −1

Answer:

(c) 1 We have:    cot150cot160cot170...cot730cot740cot750= cot150.cot160.cot170...cot450...cot730.cot740.cot750=cot150.cot160.cot170...cot450...cot(900170).cot(900160).cot(900150)=cot150.cot160.cot17...1...tan170.tan160.tan150            [cot(900θ)=tanθ and cot450=1]=1

Page No 343:

Question 9:

2sin263°+1+2sin227°3cos217°-2+3cos273°=?
(a) 32
(b) 23
(c) 2
(d) 3

Answer:

(d)3  Given: 2sin2630+1+2sin22703cos21702+3cos2730=2(sin2630+sin2270)+13(cos2170+cos2730)2=2[sin2630+sin2(900630)]+13[cos2170+cos2(900170)]2=2(sin2630+cos2630)+13(cos2170+sin2170)2        [sin(900θ)=cosθ and cos(900θ)=sinθ]=2×1+13×12                     [sin2θ+cos2θ=1]=2+132=31=3

Page No 343:

Question 10:

(sin 40° − cos 50°) = ?
(a) sin 10°
(b) cos 10°
(c) 1
(d) 0

Answer:

(d) 0We have:  (sin400cos500)=sin400cos(900400)=sin400sin400              [cos(900θ)=sinθ]=0



Page No 344:

Question 11:

(cos237° − sin253°) = ?
(a) 13
(b) 23
(c) 1
(d) 0

Answer:

(d) 0We have: (cos2370sin2530)=cos2(900530)sin2530=sin2530sin2530              [cos(900θ)=sinθ]=0

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Question 12:

(sin 43°cos 47° + cos 43°sin 47°) = ?
(a) sin 4°
(b) cos 4°
(c) 1
(d) 0

Answer:

(c) 1We have:    (sin430cos470+cos430sin470)=sin430cos(900430)+cos430sin(900430)=sin430sin430+cos430cos430              [cos(900θ)=sinθ and sin(900θ)=cosθ]=sin2430+cos2430=1

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Question 13:

sin49°cos41°-cos17°sin73°=?
(a) 1
(b) 0
(c) −1
(d) Cannot be calculated

Answer:

(b) 0We have:    (sin490cos410cos170sin730)=[sin490cos(900490)cos(900730)sin730]=(sin490sin490sin730sin730)              [cos(900θ)=sinθ ]=11=0

Page No 344:

Question 14:

(sin 79° cos 11° + cos 79° sin 11°) = ?
(a) 12
(b) 12
(c) 0
(d) 1

Answer:

(d) 1We have: (sin790cos110+cos790sin110)=sin790cos(900790)+cos790sin(900790)=sin790sin790cos790cos790             [cos(900θ)=sinθ and sin(900θ)=cosθ ]=sin2790+cos2790=1

Page No 344:

Question 15:

cot(90°-θ)·sin(90°-θ)sinθ+cot40°tan50°-(cos220°+cos270°)=?
(a) 3
(b) 2
(c) 1
(d) 0

Answer:

(c) 1We have:  [cot(900θ).sin(900θ)sinθ+cot400tan500(cos2200+cos2700)]=[tanθ.cosθsinθ+cot(900500)tan500{cos2(900700)+cos2700}]          [cot(900θ)=tanθ and sin(900θ)=cosθ]=[sinθcosθ.cosθsinθ+tan500tan500(sin2700+cos2700)]      [cos(900θ)=sinθ ]=(sinθsinθ+11)=1+11=1

Page No 344:

Question 16:

2tan230°sec252°sin238°(cosec270°-tan220°)=?
(a) 32
(b) 23
(c) 2
(d) 12

Answer:

(b) 23

We have: [2tan2300sec2520sin2380cosec2700tan2200]=[2×(13)2sec2520{sin2(900520)}{cosec2(900200)}tan2200]=[23×sec2520.cos2520sec2200tan2200]           [sin(900θ)=cosθ and cosec(900θ)=secθ]=23×11                               [sec2θtan2θ=1]=23

Page No 344:

Question 17:

(sin222°+sin268°)(cos222°+cos268°)+sin263°+cos63°sin27=?
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

(c) 2  We have:  [sin2220+sin2680cos2220+cos268+sin2630+cos630sin270]=[sin2220+sin2(900220)cos2(900680)+cos2680+sin2630+cos630{sin(900630)}]=[sin2220+cos2220sin2680+cos2680+sin2630+cos630cos630]      [sin(900θ)=cosθ and cos(900θ)=sinθ]=[11+sin2630+cos2630]                     [sin2θ+cos2θ=1]=1+1=2

Page No 344:

Question 18:

(cosec257° − tan233°) = ?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

  (b) 1We have:(cosec2570tan2330)=[cosec2(900330)tan2330]=(sec2330tan2330)                  [cosec(900θ)=secθ]=1                                                 [sec2θtan2θ=1]

Page No 344:

Question 19:

(cos228° − sin262°) = ?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

(a) 0We have:     (cos2280sin2620)=[cos2(900620)sin2620]=(sin2620sin2620)                  [cos(900θ)=sinθ]=0

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Question 20:

(sec210° − cot280°) = ?
(a) 1
(b) 0
(c) 32
(d) None of these

Answer:

(a) 1We have:   (sec2100cot2800)=[sec2100cot2(900100)]=(sec2100tan2100)                  [cot(900θ)=tanθ]=1                                                [sec2θtan2θ=1]

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Question 21:

cos (40° + θ) − sin (50° − θ) = ?
(a) 1
(b) 0
(c) sin 2θ
(d) None of these

Answer:

(b) 0We have:     [cos(400+θ)sin(500θ)]=[cos{900(500θ)}sin(500θ)]=[sin(500θ)sin(500θ)]                             [cos(900θ)=sinθ]=0

Page No 344:

Question 22:

sin (45° + θ) − cos (45° − θ) = ?
(a) 2 sin θ
(b) 2 cos θ
(c) 0
(d) 1

Answer:

(c) 0We have:     [sin(450+θ)cos(450θ)]=[sin{900(450θ)}cos(450θ)]=[cos(450θ)cos(450θ)]           [sin(900θ)=cosθ]=0



Page No 345:

Question 23:

cosec (75° + θ) −sec (15° − θ) = ?
(a) 2 sec θ
(b) 2 cosec θ
(c) 0
(d) 1

Answer:

(c)0..We have: [cosec(750+θ)sec(150θ)]=[cosec{900(150θ)}sec(150θ)]=sec(150θ)sec(150θ)                                [cosec(900θ)=secθ]=0

Page No 345:

Question 24:

If sin (θ + 34°) = cos θ and θ is acute, then θ = ?
(a) 56°
(b) 66°
(c) 28°
(d) 42°

Answer:

(d) 42°

 We have:      sin(θ+340)=cosθ=>sin(θ+340)=sin(900θ)                [cosθ=sin(900θ)]=>θ+340=900θ      =>2θ=560θ=280

Page No 345:

Question 25:

sin θ cos (90° − θ) + cos θ sin (90° − θ) = ?
(a) 0
(b) 1
(c) 2
(d) 32

Answer:

(b) 1

We have:      sinθcos(900-θ)+cosθsin(900θ)=sinθ.sinθ+cosθ.cosθ                [cos(900θ)=sinθ and sin(900θ)=cosθ]=sin2θ+cos2θ =1

Page No 345:

Question 26:

cos38° cosec52°tan18° tan35° tan60° tan72° tan55°=?
(a) 3
(b) 13
(c) 13
(d) 23

Answer:

(c) 13

We have:      [cos380cosec520tan180tan350tan600tan720tan550]=[cos380cosec(900380)tan180tan350×3×tan(900180)tan(900350)]        [cosec(900θ)=secθ and tan(900θ)=cotθ]=[cos380sec380tan180tan350×3×cot180cot350]=[1sec380×sec3801cot1801cot350×3cot180cot350]=13

Page No 345:

Question 27:

If sin 3A = cos (A − 10°), where 3A is an acute angle, then ∠A = ?
(a) 35°
(b) 25°
(c) 20°
(d) 45°

Answer:

(b) 25°We have:      [sin3A=cos(A100)]=>cos(9003A)=cos(A100)         [sinθ=cos(900θ)]=>9003A=A100=>4A=100=>A=1002541=>A=250

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Question 28:

If tan 2A = cot (A − 21°), where 2A is an acute angle, then ∠A = ?
(a) 24°
(b) 27°
(c) 35°
(d) 37°

Answer:

(d) 37°

We have:      [tan2A=cot(A210)]=>cot(9002A)=cot(A210)         [tanθ=cot(900θ)]=>9002A=A210=>3A=111=>A=1113731=>A=370

Page No 345:

Question 29:

If sec 5A = cosec (A − 30°), where 5A is an acute angle, then ∠A = ?
(a) 35°
(b) 25°
(c) 20°
(d) 27°

Answer:

(c) 20° We have:      [sec5A=cosec(A300)]=>cosec(9005A)=cosec(A300)         [secθ=cosec(900θ)]=>9005A=A300=>6A=120=>A=1202061=>A=200

Page No 345:

Question 30:

If A and B are acute angles such that sin A = cos B, then (A + B) = ?
(a) 45°
(b) 60°
(c) 90°
(d) 180°

Answer:

(c) 90°We have:      sinA=cosB=>cos(900A)=cos B         [B and   are acute]=>900A=B=>A+B=900A+B=900

Page No 345:

Question 31:

sec 70° sin 20° + cos 20° cosec 70° = ?
(a) 0
(b) 1
(c) −1
(d) 2

Answer:

(d) 2We have:      sec700sin200+cos200cosec700=sin200cos700+cos200sin700=sin200cos(900200)+cos200sin(900200)=sin200sin200+cos200cos200                        [cos(900θ)=sinθ and sin(900θ)=cosθ]=1+1=2

Page No 345:

Question 32:

cot (90° − θ) = ?
(a) cot θ
(b) −cot θ
(c) tan θ
(d) −tan θ

Answer:

(c) tan θcot(900θ)=tanθ

Page No 345:

Question 33:

If cos 9 α = sin α and 9α < 90°, then tan 5 α = ?
(a) 13
(b)  3
(c) 1
(d) 0

Answer:

(c) 1

We have:        cos9α=sinα=>cos9α=cos(900α)                [sinθ=cos(900θ)]=>9α=900α.=>10α=900=>α=90010α=90Again, tan5α=tan(5×9)0      =>tan450=1

Page No 345:

Question 34:

If cos(α + β) = 0, then sin(α − β) = ?
(a) sin α
(b) cos β
(c) sin 2α
(d) cos 2β

Answer:

(d) cos 2β

We have:        cos(α+β)=0=>cos(α+β)=cos900=>α+β=900=>α=900β         ...(i)Now, sin(αβ)=sin[(900β)β]                       [Using (i)]=sin(9002β)=cos2β                                 [sin(900θ)=cosθ]



Page No 348:

Question 1:

cos256°+cos234°sin256°+sin234°+3tan256° tan234°=?
(a) 312
(b) 4
(c) 6
(d) 5

Answer:

(b) 4

 cos2560+cos2340sin2560+sin2340+3tan2560tan2340={cos(900340)}2+cos2340{sin(900340)}2+sin2340+3{tan(900340)}2tan2340=sin2340+cos2340cos2340+sin2340+3cot2340tan2340      [cos(900θ)=sinθ, sin(900θ)=cosθ and tan(900θ)=cotθ]=11+3×1          [cotθ=1tanθ and sin2θ+cos2θ=1 ]=4

Page No 348:

Question 2:

The value of (sin230°cos245°+4tan230°+12sin290°+18cot260°)=?
(a) 38
(b) 58
(c) 6
(d) 2

Answer:

(d) 2
(sin2300cos2450)+4tan2300+12sin2900+18cot2600=122×1(2)2+4×1(3)2+12×12+18×1(3)2                [sin300=12 and cos450=12 and tan300=12 and cot600=13]=14×12+4×13+12+124=18+43+12+124=3+32+12+124=48242

Page No 348:

Question 3:

If cos A + cos2 A = 1, then (sin2 A + sin4 A) = ?
(a) 12
(b) 2
(c) 1
(d) 4

Answer:

(c) 1 cosA+2A=1=> cosA=sin2A    ...(i)Squaring both sides of (i), we get:cos2A=sin4A           ...(ii)Adding (i) and (ii), we get:sin2A+sin4A=cosA+cos2A =>sin2A+sin4A=1          [cosA+cos2A=1]

Page No 348:

Question 4:

If sin θ=32, then (cosec θ + cot θ) = ?
(a) (2+3)
(b) 23
(c) 2
(d) 3

Answer:

(d) 3
 Given: sinθ=32 and cosecθ=23cosec2θcot2θ=1=>cot2θ=cosec2θ1 =>cot2θ=431                [Given]=>cotθ=13cosecθ+cotθ=23+13 =33=3×33=3

Page No 348:

Question 5:

If cot A=45, prove that (sinA+cosA)(sinA-cosA)=9.

Answer:

Given: cotA=45Writing cot A = cos Asin A and sqauring the equation, we get:  cos2Asin2A=1625=>25cos2A=16sin2A=>25cos2A=1616cos2A=>cos2A=1641=>cosA=441sin2A=1cos2A=11641Now, sinA=2541=>sinA=541LHS=sinA+cosAsinAcosA =541+441541441=91=9=RHS

Page No 348:

Question 6:

If 2x = sec A and 2x = tan A, prove that x2-1x2=14.

Answer:

 Given: 2x=secA=>x=secA2      ...(i)and 2x=tanA=>1x=tanA2     ...(ii)x+1x=secA2+tanA2             [From (i) and (ii)] Also, x1x=secA2tanA2(x+1x)(x1x)=(secA2+tanA2)(secA2tanA2)=>x21x2=14(sec2Atan2A)x21x2=14×1                 (sec2Atan2A=1)=14 Hence proved.

Page No 348:

Question 7:

If 3 tan θ = 3 sin θ, prove that (sin2 θ − cos2 θ) = 13.

Answer:

 Given: 3tanθ=3sinθ=>3cosθ=3           [tanθ=sinθcosθ]=>cosθ=33=>cos2θ=39sin2θ=1 39=> sin2θ=69LHS=sin2θcos2θ=6939         [sin2θ=69,cos2θ=39]=39=13=RHSHence proved.



Page No 349:

Question 8:

Prove that (sin273°+sin217°)(cos228°+cos262°)=1.

Answer:

(sin273°+sin217°)(cos228°+cos262°)=1.

 LHS=sin2730+sin2170cos2280+cos2620=[sin(900170)]2+sin2170[cos(900620)]2+cos2620=cos2170+sin2170sin2620+cos2620=11                       [sin2θ+cos2θ=1]=1=RHS

Page No 349:

Question 9:

If 2 sin 2θ =3, prove that θ = 30°.

Answer:

  2sin(2θ)=3=>sin(2θ)=32=>sin(2θ)=sin(600)=>2θ=600=>θ=6002=>θ=300

Page No 349:

Question 10:

Prove that 1+cos A1-cos A = (cosec A + cot A).

Answer:

1+cos A1-cos A= (cosec A + cot A).

LHS = 1+cosA1cosAMultiplying the numerator and denominator by (1+cosA), we have:(1+cosA)2(1cosA)(1+cosA)=(1+cosA)21cos2A=1+cosAsin2A=1+cosAsinA=1sinA+cosAsinA=cosecA+cotA=RHS Hence proved.

Page No 349:

Question 11:

If cosec θ + cot θ = p, prove that cos θ = (p2-1)(p2+1).

Answer:

  cosecθ+cotθ=p=>1sinθ+cosθsinθ=p=>1+cosθsinθ=pSquaring both sides, we get:(1+cosθsinθ)2=p2=>(1+cosθ)2sin2θ=p2=>(1+cosθ)21cos2θ=p2=>(1+cosθ)2(1+cosθ)(1cosθ)=p2=>(1+cosθ)(1cosθ)=p2=>1+cosθ=p2(1cosθ)=>1+cosθ=p2p2cosθ=>cosθ(1+p2)=p21=>cosθ=p21p2+1Hence proved.

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Question 12:

Prove that (cosec A − cot A)2 = (1-cos A)(1+cos A).

Answer:

(cosec A − cot A)2 = (1-cos A)(1+cos A).

 LHS=(cosecAcotA)2=(1sinAcosAsinA)2=(1cosAsinA)2=(1cosA)2sin2A=(1cosA)21cos2A             [sin2θ+cos2θ=1]=(1cosA)(1cosA)(1cosA)(1+cosA)=(1cosA)(1+cosA)= RHSHence proved.

Page No 349:

Question 13:

If 5 cot θ = 3, find the value of 5sinθ-3cosθ4sinθ+3cosθ.

Answer:

 Given: 5cotθ=3=>5cosθsinθ=3         [cotθ=cosθsinθ]=>5cosθ=3sinθSquaring both sides, we get:25cos2θ=9sin2θ=>25cos2θ=99cos2θ            [sin2θ+cos2θ=1]=>34cos2θ=9=>cosθ=934=>cosθ=334Again, sin2θ=1cos2θ=>sin2θ=34934=2534=>sinθ=534LHS=(5sinθ3cosθ4sinθ+3cosθ)=5×5343×3344×534+3×334              [cosθ=334,sinθ=534]=25920+9=1629

Page No 349:

Question 14:

Prove that (sin 32° cos 58° + cos 32° sin 58°) = 1.

Answer:

(sin 32° cos 58° + cos 32° sin 58°) = 1

LHS=sin320cos580+cos320sin580=sin(900580)cos580+cos(900580)sin580=cos580×cos580+sin580×sin580           [sin(900θ)=cosθ,cos(900θ)=cosθ]   =cos2580+sin2580=1                     [sin2θ+cos2θ=1]=RHS

Page No 349:

Question 15:

If x = a sin θ + b cos 0 and y = a cos 0 b sin θ, prove that x2+y2=a2+b2.

Answer:

Given: x=asinθ+bcosθSquaring both sides, we get: x2=a2sin2θ+2absinθcosθ+b2cos2θ     ...(i)Also, y=acosθbsinθSquaring both sides, we get:y2=a2cos2θ2absinθcosθ+b2sin2θ    ...(ii)LHS= x2+ y2=a2sin2θ+2absinθcosθ+b2cos2θ+    a2cos2θ2absinθcosθ+b2sin2θ     [using (i)and (ii)]=a2(sin2θ+cos2θ)+b2(sin2θ+cos2θ)=a2+b2                  [sin2θ+cos2θ=1]=RHSHence proved.

Page No 349:

Question 16:

Prove that (1+sinθ)(1-sinθ) = (sec θ + tan θ)2.

Answer:

(1+sinθ)(1-sinθ)= (sec θ + tan θ)2

LHS=(1+sinθ)(1sinθ)Multiplying the numerator and denominator by (1+sinθ), we get:(1+sinθ)21sin2θ=1+2sinθ+sin2θcos2θ                   [sin2θ+cos2θ=1]=sec2θ+2×sinθcosθ×secθ+tan2θ=sec2θ+2×tanθ×secθ+tan2θ=(secθ+tanθ)2=RHSHence proved.

Page No 349:

Question 17:

Prove that 1(secθ-tanθ)-1cosθ=1cosθ-1(secθ+tanθ).

Answer:

1(secθ-tanθ)-1cosθ=1cosθ-1(secθ+tanθ)

LHS=   1secθtanθ1cosθ=(secθ+tanθ)(secθtanθ)(secθ+tanθ)secθ            Multipying the numerator and denominator by (secθ+tanθ) =secθ+tanθsec2θtan2θsecθ=secθ+tanθsecθ                   [sec2θtan2θ=1]=tanθRHS=1cosθ1secθ+tanθ=secθ(secθtanθ)sec2θtan2θ                             Multipying the numerator and denomenator by (secθtanθ)  =secθ+tanθsecθ              [sec2θtan2θ=1]=tanθLHS=RHSHence Proved

Page No 349:

Question 18:

Prove that (sin A-2sin3A)(2cos3A-cos A)=tan A.

Answer:

LHS=(sinA2sin3A)(2cos3AcosA)=sinA(12sin2A)cosA(2cos2A1)=tanA{(sin2A+cos2A2sin2A)2cos2Asin2Acos2A}              [sin2A+cos2A=1]=tanA{(cos2Asin2A)(cos2Asin2A)}=tanA=RHS

Page No 349:

Question 19:

Prove that tan A(1-cot A)+cot A(1-tan A)=(1+tan A+cot A).

Answer:

LHS=tanA(1cotA)+cotA(1tanA)=tanA(1cotA)+cot2A(cotA1)            [tanA=1cotA]=tanA(1cotA)cot2A(1cotA)=tanAcot2A(1cotA)=(1cotA)cot2A(1cotA)=1cot3AcotA(1cotA)=(1cotA)(1+cotA+cot2A)cotA(1cotA)               [a3b3=(ab)(a2+ab+b2)] =1cotA+cot2AcotA+cotAcotA=1+tanA+cotA=RHSHence proved

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Question 20:

If sec 5 A = cosec (A − 36) and 5 A is an acute angle, find the value of A.

Answer:

Given: sec5A=cosec(A360)=> cosec(9005A)=cosec(A360)         [cosec(900θ)=secθ]=> 9005A=A360=>6A=900+360 => 6A=1260=>  A=210 



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