Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 7 Trigonometric Ratios Of Complementary Angles are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios Of Complementary Angles are extremely popular among Class 10 students for Math Trigonometric Ratios Of Complementary Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 312:

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 312:

Question 2:

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 313:

Question 3:

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 313:

Question 4:

Prove that:

i sin70°cos20°+cosec20°sec70°-2cos70° cosec20°=0ii cos80°sin10°+cos59° cosec31°=2iii 2sin68°cos22°-2cot15°5tan75°-3tan45° tan20° tan40° tan50° tan70°5=1iv sin18°cos72°+3tan10° tan30° tan40° tan50° tan80°=2                   CBSE 2008v 7cos55°3sin35°-4cos70° cosec20°3tan5° tan25° tan45° tan65° tan85°=1

Answer:

i LHS=sin70°cos20°+cosec20°sec70°-2cos70° cosec20°=sin70°sin90°-20°+sec90°-20°sec70°-2cos70° sec90°-20°=sin70°sin70°+sec70°sec70°-2cos70° sec70°=1+1-2×cos70°×1cos70°=2-2=0=RHS

ii LHS=cos80°sin10°+cos59° cosec31°=cos80°cos90°-10°+sin90°-59° cosec31°=cos80°cos80°+sin31° cosec31°=1+sin31°×1sin31°=1+1=2=RHS

iii LHS=2sin68°cos22°-2cot15°5tan75°-3tan45° tan20° tan40° tan50° tan70°5=2sin68°sin90°-22°-2cot15°5cot90°-75°-3×1×cot90°-20°×cot90°-40°×tan50°×tan70°5=2sin68°sin68°-2cot15°5cot15°-3cot70° cot50° tan50° tan70°5=2-25-3×1tan70°×1tan50°×tan50°×tan70°5=2-25-35=10-2-35=55=1=RHS

iv LHS=sin18°cos72°+3tan10° tan30° tan40° tan50° tan80°=sin18°sin90°-72°+3cot90°-10°×13×cot90°-40°×tan50°×tan80°=sin18°sin18°+3cot80°×cot50°×tan50°×tan80°3=1+1tan80°×1tan50°×tan50°×tan80°=1+1=2=RHS

v LHS=7cos55°3sin35°-4cos70° cosec20°3tan5° tan25° tan45° tan65° tan85°=7cos55°3cos90°-35°-4sin90°-70° cosec20°3cot90°-5°×cot90°-25°×1×tan65°×tan85°=7cos55°3cos55°-4sin20° cosec20°3cot85° cot65° tan65° tan85°=73-4sin20°×1sin20°31tan85°×1tan65°×tan65°×tan85°=73-43=33=1=RHS

Page No 313:

Question 5:

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 313:

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 314:

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 314:

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 314:

Question 9:

If A, B and C are the angles of a ABC, prove that tanC+A2=cotB2.

Answer:

In ABC,A+B+C=180°A+C=180°-B           .....iNow,LHS=tanC+A2=tan180°-B2           Using i=tan90°-B2=cotB2=RHS

Page No 314:

Question 10:

If cos2θ=sin4θ, where 2θ and 4θ are acute angles, then find the value of θ.

Answer:

We have,cos2θ=sin4θsin90°-2θ=sin4θComparing both sides, we get90°-2θ=4θ2θ+4θ=90°6θ=90°θ=90°6 θ=15°
Hence, the value of θ is 15°.

Page No 314:

Question 11:

If sec2A = cosec(A - 42°), where 2A is an acute angle, then find the value of A.       [CBSE2008]

Answer:

We have,

sec2A=cosecA-42°cosec90°-2A=cosecA-42°Comparing both sides, we get90°-2A=A-42°2A+A=90°+42°3A=132°A=132°3 A=44°
Hence, the value of A is 44°.

Page No 314:

Question 12:

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 314:

Question 13:

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 314:

Question 14:

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 314:

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



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