Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 12 Trigonometric Ratios Of Some Complemantary Angles are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios Of Some Complemantary Angles are extremely popular among Class 10 students for Maths Trigonometric Ratios Of Some Complemantary Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 590:

Answer:

(i) sin26°cos64°


sin26°cos64°=sin90°-64°cos64°            =cos64°cos64°                  sin90°-θ=cosθ            =1Hence, sin26°cos64°=1

(ii) sec11cosec79      =sec(9079)cosec79        =cosec79cosec79         [sec 90-θ = cosec θ]=1      


(iii) tan65°cot25°


tan65°cot25°=tan90°-25°cot25°            =cot25°cot25°                  tan90°-θ=cotθ            =1Hence, tan65°cot25°=1


(iv) cos37°sin53°


cos37°sin53°=cos90°-53°sin53°            =sin53°sin53°                  cos90°-θ=sinθ            =1Hence, cos37°sin53°=1

(v)cosec42sec48    =cosec(9048)sec48        =sec48sec48        [sec 90-θ = cosec θ]    =1


(vi) cot34°tan56°



cot34°tan56°=cot90°-56°tan56°            =tan56°tan56°                  cot90°-θ=tanθ            =1Hence, cot34°tan56°=1

 

Page No 590:

Answer:

(i) LHS=cos810sin90              =cos(90090)sin90              =sin90sin90              =0              =RHS(ii) LHS=tan710cot190               =tan(900190)cot190               =cot190cot190              =0

(iii) To Prove: cosec60° − sec30° = 0


cosec60°-sec30°=cosec90°-30°-sec30° =sec30°-sec30°                             cosec90°-θ=secθ  =0Hence, cosec60°-sec30°=0.

(iv) To Prove: cot34° − tan56° = 0


cot34°-tan56°=cot90°-56°-tan56°=tan56°-tan56°                         cot90°-θ=tanθ       =0Hence, cot34°-tan56°=0.

(v) LHS=sin2480+sin2420                 =sin2(900420)+sin2420                 =cos2420+sin2420                 =1                 =RHS


(vi) To Prove: cos272° + cos218° = 1


cos272°+cos218°=cos90°-18°2+cos218°                 =sin218°+cos218°               cos90°-θ=sinθ=1                                      using the identity: cos2θ+sin2θ=1Hence, cos272°+cos218°=1.

Page No 590:

Answer:

(i) To prove: sin212° + sin2 78° = 1

sin212°+sin278°=sin90°-78°2+sin278°                 =cos278°+sin278°               sin90°-θ=cosθ=1                                      using the identity: cos2θ+sin2θ=1Hence, sin212°+sin278°=1.

(ii) To prove: sec229° – cot261° = 1

sec229°-cot261°=sec229°-cot90°-29°2                 =sec229°-tan229°               cot90°-θ=tanθ=1                                      using the identity: sec2θ-tan2θ=1Hence, sec229°-cot261°=1.


(iii) To prove: tan256° – cot234° = 0

tan256°-cot234°=tan90°-34°2-cot234°                =cot234°-cot234°               tan90°-θ=cotθ=0                                     Hence, tan256°-cot234°=0.


(iv) To prove: cos257° – sin233° = 0

cos257°-sin233°=cos90°-33°2-sin233°                 =sin233°-sin233°               cos90°-θ=sinθ=0Hence, cos257°-sin233°=0.


(v) To prove: sec250° – cot240° = 1

sec250°-cot240°=sec250°-cot90°-50°2                 =sec250°-tan250°               cot90°-θ=tanθ=1                                      using the identity: sec2θ-tan2θ=1Hence, sec250°-cot240°=1.


(vi) To prove: cosec272° – tan218° = 1

cosec272°-tan218°=cosec90°-18°2-tan218°                 =sec218°-tan218°               cosec90°-θ=secθ=1                                      using the identity: sec2θ-tan2θ=1Hence, cosec272°-tan218°=1.

Page No 590:

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) To prove: tan15° tan60° tan75° = 3

tan15° tan60° tan75°=tan15° tan60° tan90°-15°                 =tan15° tan60° cot15°               tan90°-θ=cotθ=tan15° tan60° 1tan15°             cotθ=1tanθ=tan60°=3                                         tan60°=3Hence, tan15° tan60° tan75°=3.
v LHS=tan48° tan23° tan42° tan67°=cot90°-48° cot90°-23° tan42° tan67°=cot42° cot67° tan42° tan67°=1tan42°×1tan67°×tan42°×tan67°=1=RHS

vi LHS=sin72°+cos18°sin72°-cos18°=sin72°+cos18°cos90°-72°-cos18°=sin72°+cos18°cos18°-cos18°=sin72°+cos18°0=0=RHS
(vii) To prove: cosec39° cos51° + tan21° cot69° – sec221° = 0

cosec39° cos51° +tan21° cot69°-sec221°=cosec90°-51° cos51° +tan21° cot90°-21°-sec221°                 =sec51° cos51° +tan21° tan21°-sec221°               cot90°-θ=tanθ and cosec90°-θ=secθ=sec51° cos51° +tan221°-sec221° =sec51° cos51° -sec221°-tan221°                     using the identity: sec2θ-tan2θ=1=sec51° cos51° -1=1cos51° cos51° -1                                               secθ=1cosθ=1-1=0Hence, cosec39° cos51° +tan21° cot69°-sec221°=0.

Page No 590:

Answer:

(i) sin72° + cosec72°

sin72°+cosec72°=sin90°-18°+cosec90°-18°=cos18°+sec18°                  sin90°-θ=cosθ and cosec90°-θ=secθHence, sin72°+cosec72°=cos18°+sec18°.

(ii) cosec66° + tan66°

cosec66°+tan66°=cosec90°-24°+tan90°-24°=sec24°+cot24°                  tan90°-θ=cotθ and cosec90°-θ=secθHence, cosec66°+tan66°=sec24°+cot24°.

(iii) tan68° + sec68°

tan68°+sec68°=tan90°-22°+sec90°-22°=cot22°+cosec22°                  tan90°-θ=cotθ and sec90°-θ=cosecθHence, tan68°+sec68°=cot22°+cosec22°.

(iv) cot59° + cosec59°

cot59°+cosec59°=cot90°-31°+cosec90°-31°=tan31°+sec31°                  cot90°-θ=tanθ and cosec90°-θ=secθHence, cot59°+cosec59°=tan31°+sec31°.

(v) cos51° + cot49° – sec47°

cos51°+cot49°-sec47°=cos90°-39°+cot90°-41°-sec90°-43°=sin39°+tan41°-cosec43°               cos90°-θ=sinθ, cot90°-θ=tanθ and sec90°-θ=cosecθHence, cos51°+cot49°-sec47°=sin39°+tan41°-cosec43°.

(vi) sin67° + cos75°

sin67°+cos75°=sin90°-23°+cos90°-15°=cos23°+sin15°                  sin90°-θ=cosθ and cos90°-θ=sinθHence, sin67°+cos75°=cos23°+sin15°.



Page No 591:

Answer:

Given: sin3A = cos(A – 10°)

sin3A=cosA-10°cos90°-3A=cosA-10°                   sinθ=cos90°-θ             90°-3A=A-10°90°+10°=A+3A4A=100°A=100°4A=25°Hence, A=25°.

Page No 591:

Answer:

Given: tanA = cot(A + 10°)

tanA=cotA+10°cot90°-A=cotA+10°                   tanθ=cot90°-θ             90°-A=A+10°90°-10°=A+A2A=80°A=80°2A=40°Hence, A=40°.

Page No 591:

Answer:

Given: cos2A = sin(A – 15°)

cos2A=sinA-15°sin90°-2A=sinA-15°                   cosθ=sin90°-θ             90°-2A=A-15°90°+15°=A+2A3A=105°A=105°3A=35°Hence, A=35°.

Page No 591:

Answer:

Given: sec4A = cosec(A – 15°)

sec4A=cosecA-15°cosec90°-4A=cosecA-15°                   secθ=cosec90°-θ             90°-4A=A-15°90°+15°=A+4A5A=105°A=105°5A=21°Hence, A=21°.

Page No 591:

Answer:

Given: tan2θ = cot(θ + 60°)

tan2θ=cotθ+60°cot90°-2θ=cotθ+60°                   tanθ=cot90°-θ             90°-2θ=θ+60°90°-60°=θ+2θ3θ=30°θ=30°3θ=10°Hence, the value of θ is 10°.

Page No 591:

Answer:

Given: sin(θ + 36°) = cosθ

sinθ+36°=cosθcos90°-θ+36°=cosθ                   sinθ=cos90°-θcos90°-θ-36°=cosθ54°-θ=θθ+θ=54°2θ=54°θ=54°2θ=27°Hence, θ=27°.

Page No 591:

Answer:

3cot31° tan15° cot27° tan75° cot63° cot59°=3cot90°-59° tan15° cot90°-63° tan75° cot63° cot59°=3tan59° tan15° tan63° tan75° cot63° cot59°                          cot90°-θ=tanθ=3tan59° tan90°-75° tan63° tan75° cot63° cot59°=3tan59° cot75° tan63° tan75° cot63° cot59°                          tan90°-θ=cotθ=3tan59° cot59°tan63° cot63°cot75° tan75°=3tan59° 1tan59°tan63° 1tan63°1tan75° tan75°              cotθ=1tanθ=3Hence, 3cot31° tan15° cot27° tan75° cot63° cot59°=3.

Page No 591:

Answer:

2cos58°sin32°-3cos38° cosec52°tan15° tan60° tan75°=2cos90°-32°sin32°-3cos90°-52° cosec52°tan15° tan60° tan75°=2sin32°sin32°-3sin52° cosec52°tan15° tan60° tan75°                          cos90°-θ=sinθ=21-3sin52° 1sin52°tan15° tan60° tan75°                                    cosecθ=1sinθ=2-31tan15° tan60° tan75° =2-31tan15° tan60° tan90°-15°=2-31tan15° tan60° cot15°                                         tan90°-θ=cotθ=2-31tan15° tan60° 1tan15°                                       cotθ=1tanθ=2-31tan60°=2-313                                                                  tan60°=3=2-1=1Hence, 2cos58°sin32°-3cos38° cosec52°tan15° tan60° tan75°=1.

Page No 591:

Answer:

cos70°sin20°+cos55° cosec35°tan5° tan25° tan45° tan65° tan85°=cos90°-20°sin20°+cos90°-35° cosec35°tan5° tan25° tan45° tan65° tan85°=sin20°sin20°+sin35° cosec35°tan5° tan25° tan45° tan65° tan85°                          cos90°-θ=sinθ=1+sin35° 1sin35°tan5° tan25° tan45° tan65° tan85°                                   cosecθ=1sinθ=1+1tan5° tan25° tan45° tan65° tan85° =1+1tan5° tan25° tan45° tan90°-25° tan90°-5°=1+1tan5° tan25° tan45° cot25° cot5°                                      tan90°-θ=cotθ=1+1tan5° tan25° tan45° 1tan25° 1tan5°                                  cotθ=1tanθ=1+1 tan45° =1+11                                                                                    tan45°=1=1+1=2Hence, cos70°sin20°+cos55° cosec35°tan5° tan25° tan45° tan65° tan85°=2.

Page No 591:

Answer:

sin222°+sin268°cos222°+cos268°+sin263°+cos63° sin27°=sin90°-68°2+sin268°cos222°+cos268°+sin263°+cos63° sin90°-63°=cos268°+sin268°cos222°+cos268°+sin263°+cos63° cos63°                          sin90°-θ=cosθ=cos268°+sin268°cos90°-68°2+cos268°+sin263°+cos263°=cos268°+sin268°sin268°+cos268°+sin263°+cos263°                                     cos90°-θ=sinθ=1+sin263°+cos263° =1+1                                                                                      using the identity: sin2θ+cos2θ=1=2Hence, sin222°+sin268°cos222°+cos268°+sin263°+cos63° sin27°=2.

Page No 591:

Answer:

cos65°sin25°+cosec34°sec56°-2cos43° cosec47°tan10° tan40° tan50° tan80°=cos90°-25°sin25°+cosec34°sec56°-2cos90°-47° cosec47°tan10° tan40° tan50° tan80°=sin25°sin25°+cosec34°sec56°-2sin47° cosec47°tan10° tan40° tan50° tan80°                          cos90°-θ=sinθ=1+cosec90°-56°sec56°-2sin47° cosec47°tan10° tan40° tan50° tan80°=1+sec56°sec56°-2sin47° cosec47°tan10° tan40° tan50° tan80°                                       cosec90°-θ=secθ=1+1-2sin47° 1sin47°tan10° tan40° tan50° tan80°                                                 cosecθ=1sinθ=2-2tan10° tan40° tan50° tan80°  =2-2tan10° tan40° tan90°-40° tan90°-10°=2-2tan10° tan40° cot40° cot10°                                                       tan90°-θ=cotθ=2-2tan10° tan40° 1tan40° 1tan10°                                                   cotθ=1tanθ=2-2 1 =2-2=0Hence, cos65°sin25°+cosec34°sec56°-2cos43° cosec47°tan10° tan40° tan50° tan80°=0.

Page No 591:

Answer:

cotθ tan90°-θ-sec90°-θ cosecθ+sin265°+sin225°+3 tan5° tan45° tan85°=cotθ cotθ-sec90°-θ cosecθ+sin265°+sin225°+3 tan5° tan45° tan85°                          tan90°-θ=cotθ=cot2θ- cosecθ cosecθ+sin265°+sin225°+3 tan5° tan45° tan85°                                       sec90°-θ=cosecθ=cot2θ- cosec2θ+sin90°-25°2+sin225°+3 tan5° tan45° tan85°=cot2θ- cosec2θ+cos225°+sin225°+3 tan5° tan45° tan85°                                                sin90°-θ=cosθ=-1+cos225°+sin225°+3 tan5° tan45° tan85°                                                                 using the identity: cosec2θ-cot2θ=1=-1+1+3 tan5° tan45° tan85°                                                                                         using the identity: cos2θ+sin2θ=1=0+3 tan5° tan45° tan90°-5°  =3 tan5° tan45° cot5°                                                                                                          tan90°-θ=cotθ=3 tan5° tan45° 1tan5°                                                                                                        cotθ=1tanθ=3 tan45°=3                                                                                                                                     tan45°=1Hence, cotθ tan90°-θ-sec90°-θ cosecθ+sin265°+sin225°+3 tan5° tan45°tan85°=3.

Page No 591:

Answer:

Ans

Page No 591:

Answer:

Ans

Page No 591:

Answer:

sin50°+θ-cos40°-θ+tan1° tan10° tan20° tan70° tan80° tan89°=sin50°+θ-cos40°-θ+tan1° tan10° tan20° tan90°-20° tan90°-10° tan90°-1°=sin50°+θ-cos40°-θ+tan1° tan10° tan20° cot20° cot10° cot1°                          tan90°-θ=cotθ=sin50°+θ-cos40°-θ+tan1° tan10° tan20° 1tan20° 1tan10° 1tan1°                     cotθ=1tanθ=cos90°-50°+θ-cos40°-θ+1                                                                       cos90°-θ=sinθ=cos90°-50°-θ-cos40°-θ+1=cos40°-θ-cos40°-θ+1=1Hence, sin50°+θ-cos40°-θ+tan1° tan10° tan20° tan70° tan80° tan89°=1.



Page No 593:

Answer:

sec32°cosec58°=sec90°-58°cosec58°               =cosec58°cosec58°                  sec90°-θ=cosecθ               =1Hence, the correct option is d.

Page No 593:

Answer:

cos12°sin78°+tan23°cot67°=cos90°-78°sin78°+tan23°cot67°                           =sin78°sin78°+tan23°cot67°                  cos90°-θ=sinθ                           =1+tan90°-67°cot67°                           =1+cot67°cot67°                           tan90°-θ=cotθ                           =1+1                           =2Hence, the correct option is c.

Page No 593:

Answer:

tan10° tan15° tan75° tan80°=tan90°-80° tan15° tan90°-15° tan80°=cot80° tan15° cot15° tan80°                          tan90°-θ=cotθ=1tan80°tan80°tan15° 1tan15°                  cotθ=1tanθ=1Hence, the correct option is b.

Page No 593:

Answer:

tan5° tan25° tan30° tan65° tan85°=tan90°-85° tan25° tan30° tan90°-25° tan85°=cot85° tan25°  tan30° cot25° tan85°                          tan90°-θ=cotθ=1tan85°tan85°tan25° 1tan25° tan30°                    cotθ=1tanθ=tan30°=13Hence, the correct option is c.

Page No 593:

Answer:

cos1° cos2° cos3° ...... cos180°=cos1° cos2° cos3° ...... cos90° ...... cos180°=cos1° cos2° cos3° ...... ×0× ...... cos180°                          cos90°=0=0Hence, the correct option is a.

Page No 593:

Answer:

sin43° cos47°+cos43° sin47°=sin90°-47° cos47°+cos90°-47° sin47°=cos47° cos47°+ sin47° sin47°                         sin90°-θ=cosθ and cos90°-θ=sinθ=cos247°+sin247°=1                                                                   using the identity: sin2θ+cos2θ=1Hence, the correct option is b.

Page No 593:

Answer:

sec70° sin20°+cos20° cosec70°=sec90°-20° sin20°+cos20° cosec90°-20°=cosec20° sin20°+cos20° sec20°                         sec90°-θ=cosecθ and cosec90°-θ=secθ=1sin20°sin20°+cos20°1cos20°                            cosecθ=1sinθ and secθ=1cosθ=1+1=2                                                                   Hence, the correct option is c.

Page No 593:

Answer:

cosec257°-tan233°=cosec90°-33°2 -tan233°=sec233°-tan233°                         cosec90°-θ=secθ=1                                                using the identity:sec2θ-tan2θ=1                                                                  Hence, the correct option is a.

Page No 593:

Answer:

sec210°-cot280°=sec90°-80°2 -cot280°=cosec280°-cot280°                      sec90°-θ=cosecθ=1                                                using the identity:cosec2θ-cot2θ=1Hence, the correct option is b.

Page No 593:

Answer:

2sin263°+1+2sin227°3cos217°-2+3cos273°=2sin90°-27°2+1+2sin227°3cos90°-73°2-2+3cos273°=2cos227°+1+2sin227°3sin273°-2+3cos273°                      sin90°-θ=cosθ and cos90°-θ=sinθ=2cos227°+sin227°+13sin273°+cos273°-2=21+131-2                                            using the identity:sin2θ+cos2θ=1=2+13-2=3Hence, the correct option is d.



Page No 594:

Answer:

sin38°-cos52°=sin90°-52°-cos52°=cos52°-cos52°                      sin90°-θ=cosθ=0Hence, the correct option is a.

Page No 594:

Answer:

sin223°+sin267°cos213°+cos277°+sin259°+cos59° sin31°=sin223°+sin90°-23°2cos213°+cos90°-13°2+sin259°+cos59° sin90°-59°=sin223°+cos223°cos213°+sin213°+sin259°+cos59° cos59°                      sin90°-θ=cosθ and cos90°-θ=sinθ=11+sin259°+cos259°                                                      using the identity: sin2θ+cos2θ=1=1+1=2 Hence, the correct option is b.


Disclaimer: There must be sin31° instead of sin21°.

Page No 594:

Answer:

2tan230° sec252° sin238°cosec270°-tan220°=2tan230° sec90°-38°2 sin238°cosec90°-20°2-tan220°=2tan230° cosec238° sin238°sec220°-tan220°                      sec90°-θ=cosecθ and cosec90°-θ=secθ=2tan230° cosec238° sin238°1                     using the identity: sec2θ-tan2θ=1=2tan230° 1sin238° sin238°                             cosecθ=1sinθ=2tan230°=2132                                                    tan30°=13=23Hence, the correct option is c.

Page No 594:

Answer:

cos38° cosec52°tan18° tan35° tan60° tan72° tan55°=cos38° cosec52°tan90°-72° tan90°-55° tan60° tan72° tan55°=cos38° cosec90°-38°cot72° cot55° tan60° tan72° tan55°                          tan90°-θ=cotθ=cos38° sec38°cot72° cot55° tan60° tan72° tan55°                          cosec90°-θ=secθ=cos38° 1cos38°1tan72° 1tan55° tan60° tan72° tan55°                      secθ=1cosθ and cotθ=1tanθ=1 tan60°=13                                                                          tan60°=3Hence, the correct option is b.

Page No 594:

Answer:

sin60°+θ-cos30°-θ=cos90°-60°+θ-cos30°-θ            sinθ=cos90°-θ=cos90°-60°-θ-cos30°-θ=cos30°-θ-cos30°-θ=0Hence, the correct option is c.

Page No 594:

Answer:

Given: sinA = cosB

sinA=cosBcos90°-A=cosB            sinθ=cos90°-θ90°-A=B90°=B+AA+B=90°Hence, the correct option is d.

Page No 594:

Answer:

Given: cos(α + β) = 0

As we know that,cos90°=0Since, cosα+β=0α+β=90°α=90°-β            ...1Now,sinα-β=sin90°-β-β              =sin90°-2β              =cos2β                       sin90°-θ=cosθHence, the correct option is b.

Page No 594:

Answer:

sin45°+θ-cos45°-θ=cos90°-45°+θ-cos45°-θ            sinθ=cos90°-θ=cos90°-45°-θ-cos45°-θ=cos45°-θ-cos45°-θ=0Hence, the correct option is a.

Page No 594:

Answer:

Given: sec4A = cosec(– 10°)

sec4A=cosecA-10°cosec90°-4A=cosecA-10°                   secθ=cosec90°-θ             90°-4A=A-10°90°+10°=A+4A5A=100°A=100°5A=20°Hence, the correct option is a.

Page No 594:

Answer:

Given: sin3A = cos(A – 10°)

sin3A=cosA-10°cos90°-3A=cosA-10°                   sinθ=cos90°-θ             90°-3A=A-10°90°+10°=A+3A4A=100°A=100°4A=25°Hence, the correct option is c.

Page No 594:

Answer:

(i) cot34° – tan56° = .......

cot34°-tan56°=cot90°-56°-tan56°=tan56°-tan56°                  cot90°-θ=tanθ=0Hence, cot34°-tan56°=0.


(ii) cosec31° – sec59° = .......

cosec31°-sec59°=cosec90°-59°-sec59°=sec59°-sec59°                  cosec90°-θ=secθ=0Hence, cosec31°-sec59°=0.

(iii) cos267° + cos223° = ..........

cos267°+cos223°=cos90°-23°2+cos223°                 cos90°-θ=sinθ=sin223°+cos223°                               sin2θ+cos2θ=1=1Hence, cos267°+cos223°=1.

(iv) cosec254° – tan236° = .........

cosec254°-tan236°=cosec90°-36°2-tan236°                 cosec90°-θ=secθ=sec236°-tan236°                                  sec2θ-tan2θ=1=1Hence, cosec254°-tan236°=1.

(v) sec240° – cot250° = ..........

sec240°-cot250°=sec90°-50°2-cot250°                 sec90°-θ=cosecθ=cosec250°-cot250°                           cosec2θ-cot2θ=1=1Hence, sec240°-cot250°=1.



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