RS Aggarwal Solutions Class 10 Chapter 7 Trigonometric Identities


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RS Aggarwal Solutions Class 10 Chapter 7 Trigonometric Identities





RS Aggarwal Class 10 Solutions Trigonometric Identities Exercise 7A





Question 1:
Solution:





(i) (1 – cos2θ) cosec2θ = 1





L.H.S. = (1 – cos2θ) cosec2θ





= (sin2θ) × cosec2θ





(Using identity sin2θ + cos2 θ = 1)





= 1/ cosec2θ × cosec2θ





= 1





= R.H.S.





Hence Proved.





(ii) (1 + cot2θ) sin2θ = 1





L.H.S. = (1 + cot2θ) × sin2 θ





= (cosec2 θ) × sin2 θ





(Using identity 1 + cot2 θ = cosec2 θ)





= 1/ sin2θ × sin2 θ





= 1





= R.H.S.





Hence Proved.





Question 2:





Solution:





(i) (sec2θ − 1) cot2θ = 1





L.H.S. = (sec2 θ – 1) × cot2 θ





= (tan2θ) x cot2θ





(using identity 1 + tan2 θ = sec2 θ)





= 1/cot2θ x cot2θ





= 1





= R.H.S.





Hence Proved.





(ii) (sec2θ − 1) (cosec2θ − 1) = 1





L.H.S. = (sec2 θ – 1)(cosec2 θ – 1)





= (tan2θ) × cot2θ





(using identity 1 + cot2 θ = cosec2 θ and 1 + tan2 θ = sec2 θ)





= tan2θ x 1/tan2θ





= 1





= R.H.S.





Hence Proved.





(iii) (1− cos2θ) sec2θ = tan2θ





L.H.S. = (1 – cos2 θ) sec2 θ





= (sin2θ) × (1/cos2θ)





(using identity sin2 θ = 1- cos2 θ)





= tan2 θ





= R.H.S.





Hence Proved.





Question 3: Prove





rs aggarwal class 10 chapter 13a 1




Question 4: Prove
Solution:





(i)(1 + cos θ) (1 − cos θ) (1 + cot2θ) = 1





LHS: (1 + cos θ) (1 − cos θ) (1 + cot2θ)





= (1 – cos2 θ) × cosec2 θ





(Using sin2 θ + cos2 θ = 1)





= (sin2 θ) × cosec2 θ





= sin2 θ x 1/sin2 θ





= 1





= R.H.S.





Hence Proved





(ii) cosec θ (1 + cos θ) (cosec θ − cot θ) = 1





L.H.S.





cosec θ (1 + cos θ) (cosec θ − cot θ)





= (cosec θ + cosec θ cos θ)(cosec θ – cot θ)





We know, cosec θ = 1/sin θ and cot θ = cosθ/sinθ





= (cosec θ + cot θ)(cosec θ – cotθ)





Apply formula: (a + b)(a – b) = a2 – b2





= cosec2 θ – cot2 θ





= 1





= R.H.S.





Hence proved.





Question 5:
Solution:





(i)





L.H.S.





= cot2 θ – 1/sin2θ





= cos2θ/sin2θ – 1/sin2θ





= (cos2θ – 1)/sin2θ





=-sin2θ/sin2θ





= -1





= R.H.S





(ii)





L.H.S.





= tan2 θ – 1/cos2θ





= sin2θ/cos2θ – 1/cos2θ





= (sin2θ – 1)/cos2θ





=-cos2θ/cos2θ





= -1





= R.H.S





(iii)





L.H.S.





= cos2 θ + 1/(1+cot2θ





= cos2 θ + 1/cose2θ





= cos2 θ + sin2θ





= 1





= R.H.S





Question 6: Prove





rs aggarwal class 10 chapter 13a 3




Question 7:





Solution:





(i) L.H.S.





= sec θ (1 − sin θ) (sec θ + tan θ)





Examples on Trigonometric Identities




= cos2θ/cos2θ





= 1





= R.H.S.





(ii) L.H.S. = sin θ (1 + tan θ) + cos θ (1+ cot θ)





= sin θ (1 + sin θ/cos θ) + cos θ (1+ cos θ/sinθ)





= sin θ{(cosθ +sinθ )/cos θ} + cos θ{(sinθ+cosθ )/sinθ}





=(cos θ + sin θ) (sinθ/cos θ + cos θ/sinθ)





= (cos θ + sin θ)/cos θ sin θ





= cosec θ + sec θ





= R.H.S.





Question 8:





rs aggarwal class 10 chapter 13a 5




(ii)





rs aggarwal class 10 chapter 13a 6




= 1/cos θ





= sec θ





= R.H.S.





Question 9: Prove





rs aggarwal class 10 chapter 13a 7




Question 10: Prove





rs aggarwal class 10 chapter 13a 8









Exercise 7B





Question 1:
Solution:





a cos θ + b sin θ = m





Squaring equation, we get





a2 cos2 θ + b2 sin2 θ + 2ab cos θ sin θ = m2 ……(1)





Again Square equation, a sin θ – b cos θ = n





a2 sin2 θ + b2 cos2 θ – 2ab cos θ sin θ = n2 ……(2)





Add (1) and (2)





a2 cos2 θ + b2 sin2 θ + 2ab cos θ sin θ + a2 sin2 θ + b2 cos2 θ – 2ab cos θ sin θ = m2 +n2





a2 (cos2 θ + sin2θ) + b2 (cos2 θ + sin2 θ) = m2 + n2





(Using cos2 θ + sin2θ = 1)





a2 + b2 = m2 + n2





Hence Proved.





Question 2:
Solution:





a sec θ + b tan θ = x





a tan θ + b sec θ = y





Squaring above equations:





a2 sec2 θ + b2 tan2 θ + 2ab sec θ tan θ = x2 …..(1)





a2 tan2 θ + b2 sec2 θ + 2ab sec θ tan θ = y2 ….(2)





Subtract equation (2) from (1):





a2 (sec2 θ – tan2 θ) + b2 (tan2 θ – sec2 θ) = x2 – y2





(using sec2 θ = 1 + tan2 θ)





or a2 – b2 = x2 – y2





Hence proved.





Question 3:
Solution:





x/a sin θ – y/b cos θ = 1





x/a cos θ + y/b sin θ = 1





Squaring both the equations, we have





x2/a2 sin2 θ + y2/b2 cos2 θ – 2 xy/ab cos θ sin θ = 1 ….(1)





x2/a2 cos2 θ + y2/b2 sin2 θ + 2 xy/ab cos θ sin θ = 1 ……(2)





Add (1) and (2), we get





x2/a2(sin2 θ + cos2 θ) + y2/b2 (sin2 θ + cos2 θ) = 1+1





(Using cos2 θ + sin2θ = 1)





x2/a2 + y2/b2 = 2





Question 4:
Solution:





(sec θ + tan θ) = m …(1) and





(sec θ − tan θ) = n ….(2)





Multiply (1) and (2), we have





(sec θ + tan θ) (sec θ – tan θ) = mn





(sec2 θ – tan2 θ) = mn





(Because sec2 θ – tan2 θ=1)





1 = mn





Or mn = 1





Hence Proved





Question 5:





Solution:





(cosec θ + cot θ) = m …(1) and





(cosec θ − cot θ) = n …(2)





Multiply (1) and (2)





(cosec2 θ – cot2 θ) = mn





(Because cosec2 θ – cot2 θ = 1)





1 = mn





Or mn = 1





Hence Proved





Question 6:
Solution:





x = a cos3 θ





y = b sin3 θ





L.H.S.





rs aggarwal class 10 chapter 13b 3




= cos2 θ + sin2 θ





= 1





= R.H.S.





Question 7:
Solution:





(tan θ + sin θ) = m and (tan θ − sin θ) = n





To Prove: (m2 − n2)2 = 16mn





L.H.S. = (m2 − n2)2





= [(tan θ + sin θ)2 – (tan θ − sin θ)2]2





= (4tan θ sin θ)2





= 16 tan2 θ sin2 θ …(1)





R.H.S. = 16mn





= 16(tan θ + sin θ)(tan θ − sin θ)





= 16(tan2 θ − sin2 θ)





= 16 [{sin2 θ (1-cos2 θ)/cos2θ]





= 16 x sin2 θ/cos2θ x (1-cos2 θ)





= 16 tan2 θ sin2 θ …(2)





From (1) and (2)





L.H.S. = R.H.S.





Question 8:
Solution:





(cot θ + tan θ) = m and (sec θ − cos θ) = n





m = 1/tanθ + tan θ = (1+ tan2 θ)/ tan θ = sec2 θ / tan θ





= 1/sinθcosθ





or m = 1/sinθcosθ





Again, n = sec θ − cos θ





= 1/cosθ − cos θ





= (1 – cos2 θ)/cosθ





= sin2 θ/cos θ





or n = sin2 θ/cos θ





To prove: (m2n)^(2/3) − (mn2)^(2/3) = 1





L.H.S.





(m2n)^(2/3) − (mn2)^(2/3)





Substituting the values of m and n, we have





rs aggarwal class 10 chapter 13b 4




= (1 – sin2 θ)cos2 θ





(We know, 1 – sin2 θ = cos2 θ)





= cos2 θ/ co2 θ





= 1





=R.H.S.





Hence proved.





Question 9:
Solution:





(cosec θ − sin θ) = a3 and (sec θ − cos θ) = b3





(cosec θ − sin θ) = a3





(1/sinθ − sin θ) = a3





cos2θ/sinθ = a3





And a2 = (a3)^(2/3) = (cos2θ/sinθ )^(2/3) …..(1)





Again





(sec θ − cos θ) = b3





(1/cosθ − cos θ) = b3





= sin2 θ/cosθ = b3





And, b2 = (b3)^(2/3) = (sin2 θ/cosθ)^(2/3)





To Prove:a2b2(a2 + b2) = 1





L.H.S.





a2b2(a2 + b2)





rs aggarwal class 10 chapter 13b 5




= sin2 θ + cos2 θ





= 1





=R.H.S.





Hence proved.





Question 10:
Solution:





(2 sin θ + 3 cos θ) = 2 …(1)





(2 sin θ + 3 cos θ)2 + (3 sin θ – 2 cos θ)2





= 4sin2 θ + 9 cos2 θ + 12sin θ cos θ + 9 sin2 θ + 4 cos2 θ – 12 sin θ cos θ





= 13sin2 θ + 13 cos2 θ





= 13(sin2 θ + cos2 θ)





= 13





(Because (sin2 θ + cos2 θ) = 1)





=> (2 sin θ + 3 cos θ)2 + (3 sin θ – 2 cos θ)2 = 13





Using equation (1)





=> (2)2 + (3 sin θ – 2 cos θ)2 = 13





=> (3 sin θ – 2 cos θ)2 = 9





or (3 sin θ – 2 cos θ) = ± 3





Hence Proved.





Exercise 7C





Question 1:
Solution:





(1 – sin2θ) sec2 θ = (cos2 θ) × 1/cos2 θ





= 1





Question 2:
Solution:





(1-cos2θ)cosec2θ





= sin2θ x 1/sin2θ





= 1





Question 3:
Solution:





(1+tan2θ)cos2θ = sec2 θ x 1/sec2 θ





= 1





Question 4:
Solution:





(1+cot2θ)sin2θ = cose2θ x 1/cose2θ





= 1





Question 5:
Solution:





sin2θ + 1/(1+tan2θ)





= sin2θ + 1/(sec2θ)





= sin2θ + cos2θ





= 1





Question 6:
Solution:





(cot2θ – 1/sin2θ ) = (cos2θ/sin2θ – 1/sin2θ )





= (cos2θ – 1)/sin2θ





= -sin2θ/sin2θ





= -1





Question 7:
Solution:





sinθ cos(90°-θ)+cosθ sin(90°-θ) = sinθ x sinθ + cosθ x cosθ





= sin2θ + cos2θ





= 1





Question 8:
Solution:





cosec2(90°-θ) – tan2θ = sec2θ – tan2θ





= 1





Question 9:
Solution:





sec2θ(1+sinθ)(1-sinθ) = sec2θ(1-sin2 θ)





= sec2θ x cos2θ





= 1/cos2θ x cos2θ





= 1





Question 10:
Solution:





cosec2θ(1+cosθ)(1-cosθ) = cosec2θ (1-cos2 θ)





= cosec2θ x sin2θ





= cosec2θ x 1/cosec2θ





= 1





Question 11:
Solution:





sin2θ cos2θ(1+tan2θ)(1+cot2θ) = sin2θ x cos2θ x sec2θ x cosec2θ





= sin2θ x cos2θ x 1/cos2θ x 1/sin2θ





=1





Question 12:
Solution:





(1+tan2θ)(1+sinθ)(1-sinθ) = sec2θ(1-sin2 θ)





= sec2θ x cos2θ





= 1/cos2θ x cos2θ





= 1





Question 13:
Solution:





3cot2θ – 3cosec2θ = 3(cot2θ – cosec2θ)





= 3 x -1





= -3





Question 14:
Solution:





4tan2θ – 4/cos2θ = 4 x sin2θ/cos2θ – 4/cos2θ





= (4(sin2θ – 1))/cos2θ





= 4 (-cos2θ) / cos2θ





= -4





Question 15:
Solution:





(tan2θ – sec2θ) / (cot2θ – cose2θ) = -1/-1





= 1





(Using 1 + cot2θ = cose2θ and 1 + tan2θ = sec2θ)





Complete RS Aggarwal Solutions Class 10





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