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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
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
Question 7:
Solution:
(i) L.H.S.
= sec θ (1 − sin θ) (sec θ + tan θ)
= 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:
(ii)
= 1/cos θ
= sec θ
= R.H.S.
Question 9: Prove
Question 10: Prove
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.
= 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
= (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)
= 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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