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Page No 590:

Question 1:

Without using trigonometric tables, evaluate:
(i) sin 26°cos 64°

(ii) sec 11°cosec 79°

(iii) tan 65°cot 25°

(iv) cos 37°sin 53°

(v) cosec 42°sec 48°

(vi) cot 34°tan 56°

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:

Question 2:

Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 60° − sec 30° = 0
(iv) cot 34° − tan 56° = 0
(v) sin248° + sin242° = 1
(vi) cos272° + cos218° = 1

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:

Question 3:

Without using trigonometric tables, prove that:
(i) sin212° + sin2 78° = 1
(ii) sec229° – cot261° = 1
(iii) tan256° – cot234° = 0
(iv) cos257° – sin233° = 0
(v) sec250° – cot240° = 1
(vi) cosec272° – tan218° = 1

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:

Question 4:

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) tan 15° tan 60° tan 75° = 3
(v) tan48° tan23° tan42° tan67° tan 45° = 1
(vi) (sin72° + cos18°)(sin72° − cos18°) = 0
(vii) cosec 39° cos 51° + tan 21° cot 69° – sec221° = 0

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:

Question 5:

Express each of the following in terms of trigonometric ratios of angles between 0° and 45°.
(i) sin 72° + cosec 72°
(ii) cosec 66° + tan 66°
(iii) tan 68° + sec 68°
(iv) cot 59° + cosec 59°
(v) cos 51° + cot 49° – sec 47°
(vi) sin 67° + cos 75°

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:

Question 6:

If sin 3A = cos (A – 10°), where 3A is an acute angle then find ∠A.

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:

Question 7:

It tan A = cot (A + 10°), where (A + 10°) is acute then find ∠A.

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:

Question 8:

If cos 2A = sin (A – 15°), where 2A is acute then find ∠A.

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:

Question 9:

If sec 4A = cosec (A – 15°), where 4A is acute then find ∠A.

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:

Question 10:

If tan 2θ = cot (θ + 60°), where 2θ and (θ + 6°) an acute angles, find the value of θ.

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:

Question 11:

If sin (θ + 36°) = cos θ, where (θ + 36°) is acute, find θ.

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:

Question 12:

3cot 31° tan 15° cot 27° tan 75° cot 63° cot 59°

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:

Question 13:

2cos 58°sin 32°-3cos 38° cosec 52°tan 15° tan 60° tan 75°

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:

Question 14:

cos 70°sin 20°+cos 55° cosec 35°tan 5° tan 25° tan 45° tan 65° tan 85°

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:

Question 15:

sin222°+sin268°cos222°+cos268°+sin263°+cos63° sin 27°

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:

Question 16:

cos 65°sin 25°+cosec 34°sec 56°-2cos 43° cosec 47°tan 10° tan 40° tan 50° tan 80°

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:

Question 17:

cot θ tan 90°-θ-sec90°-θcosec θ+sin265°+sin225°+3 tan 5°tan 45°tan 85°

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:

Question 18:

sec90°-θ cosecθ-tan90°-θ cotθ+cos225°+cos265°3tan27° tan63°

Answer:

Ans

Page No 591:

Question 19:

cos(90°-θ)sec(90°-θ)tanθcosec(90°-θ)sin(90°-θ)cot(90°-θ)+tan(90°-θ)cotθ

Answer:

Ans

Page No 591:

Question 20:

sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°

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:

Question 1:

sec 32°cosec 58°=?

(a) 23

(b) 1629

(c) 163

(d) 1

Answer:

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

Page No 593:

Question 2:

cos 12°sin 78°+tan 23°cot 67°=?
(a) 0
(b) 1
(c) 2
(d) 32

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:

Question 3:

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

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:

Question 4:

tan 5° tan 25° tan 30° tan 65° tan 85° = ?
(a) 1
(b) 3
(c) 13
(d) 12

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:

Question 5:

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

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:

Question 6:

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

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:

Question 7:

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

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.

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Question 8:

cosec257° – tan233° = ?
(a) 1
(b) 0
(c) –1
(d) 2

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.

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Question 9:

sec210° – cot280° = ?
(a) 0

(b) 1

(c) 34

(d) 12

Answer:

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

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Question 10:

2sin263°+1+2sin227°3cos217°-2+3cos273°=?

(a) 23

(b) 32

(c) 2

(d) 3

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.



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Question 11:

sin 38° – cos 52° = ?
(a) 0

(b) 1

(c) 32

(d) 23

Answer:

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

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Question 12:

sin223°+sin267°cos213°+cos277°+sin259°+cos59°sin 21°=?
(a) 3
(b) 2
(c) 1
(d) 0

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°.

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Question 13:

2tan230° sec252° sin238°cosec270°-tan220°=?

(a) 2

(b) 1

(c) 23

(d) 32

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.

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Question 14:

cos 38° cosec 52°tan 18° tan 35° tan 60° tan 72° tan 55°=?

(a) 13

(b) 13

(c) 3

(d) 23

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.

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Question 15:

sin (60° + θ) – cos (30° – θ) = ?
(a) 2sin θ
(b) 2cos θ
(c) 0
(d) 1

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.

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Question 16:

If A and B are acute angles such that sin A = cos B then (A + B) = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

Given: sinA = cosB

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

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Question 17:

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

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.

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Question 18:

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

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.

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Question 19:

If sec 4A = cosec (A – 10°) and 4A is acute then A = ?
(a) 20°
(b) 25°
(c) 30°
(d) 40°

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.

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Question 20:

If sin 3A = cos (A – 10°) and 3A is acute then A = ?
(a) 15°
(b) 20°
(c) 25°
(d) 30°

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.

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Question 21:

Fill in the blanks.
(i) cot 34° – tan 56° = .......
(ii) cosec 31° – sec 59° = .......
(iii) cos267° + cos223° = ..........
(iv) cosec254° – tan236° = .........
(v) sec240° – cot250° = ..........

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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