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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