RS Aggarwal Solutions Class 10 Chapter 8 Trigonometric Ratios of Complementary Angles


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RS Aggarwal Solutions Class 10 Chapter 8 Trigonometric Ratios of Complementary Angles





Exercise 8





Question 1.





rs aggarwal class 10 chapter 12 img 1




rs aggarwal class 10 chapter 12 img 2




Question 2:





Solution:





(i) LHS = cos81° – sin9°





= cos(90° -9°)- sin9°





= sin9° – sin9°





= 0





= RHS





(ii) LHS = tan71° – cot19°





=tan(90° – 19°) – cot19°





=cot19° – cot19°





=0





= RHS





(iii) LHS = cosec80° – sec10°





= cosec(90° – 10°) – sec(10°)





= sec10° – sec10°





= 0





= RHS





(iv) cosec^2 72° − tan^2 18° = 1





LHS = cosec^2 72° – tan^2 18°





= cosec^2 72° – tan2 (90 – 72)°





= cosec^2 72° – cot^2 72°





= 1 = RHS





(v) cos^2 75° + cos^2 15° = 1





LHS = cos^2 75° + cos^2 15°





= cos^2 75° + cos^2 (90 – 75)°





= cos^2 75° + sin^2 75°





= 1= RHS





(vi) tan^2 66° − cot^2 24° = 0





LHS = tan^2 66° − cot^2 24°





= tan^2 66° – cot^2 (90 – 66)°





= tan^2 66° – tan^2 66°





= 0





= RHS





(vii) sin^2 48° + sin^2 42° = 1





LHS = sin^2 48° + sin^2 42°





= sin^2 48° + sin^2 (90 – 48)°





= sin^2 48° + cos^2 48°





= 1





= RHS





(viii) cos^2 57° − sin^2 33° = 0





LHS = cos^2 57° − sin^2 33°





= cos^2 57° – sin^2 (90 – 57)°





= cos^2 57° – cos^2 57°





= 0





=RHS





(ix) (sin 65° + cos 25°)(sin 65° − cos 25°) = 0





LHS = (sin 65° + cos 25°)(sin 65° − cos 25°)





= sin^2 65° – cos^2 25°





= sin^2 65° – cos^2 (90 – 65)°





= sin^2 65° – sin^2 65°





= 0





=RHS





Question 3.
Solution:





(i) LHS = sin53° cos37° + cos53° sin37°





= sin53° cos(90° – 53°) + cos53° sin (90° – 53°)





= sin53° x sin53° + cos53° x cos53°





= sin^2 53° + cos^2 53°





= 1





= RHS





(ii) LHS = cos54° cos36° − sin54° sin36°





= cos54° cos36° − sin(90° -36°) sin(90° -54°)





= cos54° cos36° – cos36°cos54°





= 0





=RHS





(iii) LHS = sec70° sin20° + cos20° cosec70°





= sec(90° – 20°) sin20° + cos20° cosec(90° – 20°)





= cosec 20° sin20° + cos20° sec 20°





= 1 +1





= 2





=RHS





(iv) LHS = sin35° sin55° − cos35° cos55°





= sin(90° – 55°) sin(90° – 35°) − cos35° cos55°





= cos55° cos35° – cos35° cos55°





= 0





=RHS





(v) LHS = (sin72° + cos18°)(sin72° − cos18°)





=(sin^2 72° – cos^2 18°)





= (sin^2 72° – cos^2 (90° – 72°))





= sin^2 72° – sin^2 72°





= 0





=RHS





(vi) LHS = tan48° tan23° tan42° tan67°





=tan48° tan23° tan (90° – 48°) tan (90° – 23°)





= tan48° tan23° cot48° cot23°





= 1×1





= 1





=RHS





Question 4.





Solution:





rs aggarwal class 10 chapter 12 img 4




rs aggarwal class 10 chapter 12 img 5




(v)





rs aggarwal class 10 chapter 12 img 6




Question 5.
Solution:





(i)





LHS = sin θ cos (90° – θ) + sin (90° – θ) cos θ





= sin θ sin θ + cos θ cos θ





= sin^2 θ + cos^2 θ





= 1





= R.H.S.





Hence proved.





rs aggarwal class 10 chapter 12 img 8




rs aggarwal class 10 chapter 12 img 9




(vi)





rs aggarwal class 10 chapter 12 img 10




=(1+1)/(3×1)





= 2/3 = RHS





Hence proved.





(vii)





LHS = cot θ tan (90° -θ) – sec (90° – θ)cosec θ +√3tan 12°tan 60°tan 78°





= cot θ cot θ – cosec θ cosec θ +√3 tan 60° tan 12° tan 78°





= cot^2 θ – cosec^2 θ +√3 tan 60° tan 12° tan(90-12)°





= – (cosec^2 θ – cot^2 θ) +√3 tan 60° tan 12° cot 12°





= – 1 + √3(√3 × 1)





= – 1 + 3





= 2





= R.H.S.





Hence proved.





Question 6:
Solution:





rs aggarwal class 10 chapter 12 img 11




rs aggarwal class 10 chapter 12 img 12




Question 7:
Solution:





L.H.S.





= sin (70°+θ) — cos (20° — θ)





= sin (70°+θ) — cos [90°-(70° + θ)]





= sin (70°+θ) — sin (70° + θ)





= 0





= R.H.S.





Hence Proved.





(ii)





L.H.S.





= tan (55° — θ) — cot (35° + θ)





= tan (90°-(35° +θ)) — cot (35° + θ)





= cot (35° + θ) – cot (35° + θ)





= 0





=RHS





Hence Proved.





(iii)





L.H.S.





= cosec (67° + θ) — sec (23° — θ)





= cosec (67° + θ) – sec (90° —(23° + θ))





= cosec (67° + θ) – cosec (67° + θ)





= 0





=RHS





Hence Proved.





(iv)





L.H.S.





= cosec (65° + θ) — sec (25° — θ) — tan (55° — θ) + cot (35° + θ)





= cosec (65° + θ) — sec (90° – (65° + θ)) — tan (90° – (35° + θ)) + cot (35° + θ)





= cosec (65° + θ) — cosec (65° + θ)) — cot (35° + θ) + cot (35° + θ)





= 0





= R.H.S.





(v)





L.H.S.





= sin (50° +θ) — cos (40° — θ) + tan 1° tan 10° tan 80° tan 89°





= sin ((90°-(40° – θ)) — cos (40° — θ) + (tan 1° tan 89°)(tan10° tan 80°)





= cos (40° – θ) – cos (40° – θ) + {tan 1° tan (90°-1°)}{tan10° tan(90°-10°)}





= 0 + {(tan 1° cot 1°}{tan10° cot 10°}





= 0 + 1 = 1





= R.H.S.





Question 8.





Solution:





rs aggarwal class 10 chapter 12 img 13




Question 9.
Solution:





Given function is : tanC+A2=cotB2





Sum of all the angles of a triangle = 180 degree





So, A + B + C = 180o





Or A + C = 180o – B





And, (A + C)/2 = (180o – B)/2 = 90o – B/2





Now, tan (A + C)/2 = tan(90o – B/2) = cot B/2





Hence Proved.





Question 10.
Solution:





cos 2θ = sin 4θ …(1)





We know that,





sin( 90° – θ) = cos θ





So, equation (1) can be written as





sin (90° – 2θ) = sin 4θ





On comparing both sides





90° – 2θ = 4θ





90° = 4θ + 2θ





6θ = 90°





or θ = 15°





The value of θ is 15°





Question 11:
Solution: Given: sec2A = cosec(A − 42°)





rs aggarwal class 10 chapter 12 img 14




The value of angle A is 44 degrees.





Question 12:
Solution:





sin 3 A = cos (A − 26°) (given)





or cos (90° – 3A) = cos (A – 26°)





On comparing





90° – 3A = A – 26°





A + 3A = 90° + 26°





4A = 116° = 29°





The value of A is 29°.





Question 13:
Solution:





tan 2A = cot (A – 12°)





or cot (90° – 2A) = cot (A – 12°)





On comparing





90° – 2A = A – 12 °





A + 2A = 90° + 12°





3A = 102°





A = 34°





The value of A is 34°





Question 14:
Solution: sec 4A = cosec (A – 15°)





or cosec (90° – 4A) = cosec (A – 15°)





On comparing





90° – 4A = A – 15°





A + 4A = 90° + 15°





5A = 105°





A = 21°





The value of A is 21°.





Question 15:





Solution:





= 2/3 cosec^2 58°- 2/3 cot 58° tan 32° – 5/3 tan 13° tan 37° tan 45° tan 53° tan 77°





= 2/3 cosec^2 58°- 2/3 cot 58° tan (90-58)° – 5/3 tan 45° (tan 13° tan 77°) (tan 37° tan 53°)





= 2/3 cosec^2 58°- 2/3 cot 58° cot 58° – 5/3 × 1 x (tan 13° tan (90-13)°) × (tan 37° tan (90-37)°)





= 2/3 cosec^2 58°- 2/3 cot^2 58° – 5/3 × (tan 13° cot 13°) (tan 37° cot 37°)





= 2/3 [cosec^2 58°- cot^2 58°] – 5/3





= (2/3) – (5/3)





= -1





= R.H.S.





Complete RS Aggarwal Solutions Class 10





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RS Aggarwal Solutions Class 10 Chapter 8 Trigonometric Ratios of Complementary Angles RS Aggarwal Solutions Class 10 Chapter 8 Trigonometric Ratios of
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