R S Aggarwal Solutions for Class 10 Maths Chapter 13 Trigonometric Identities Exercise 13B

R S Aggarwal Solutions Class 10 for maths are designed by subject experts at BYJU’S after extensive research to provide best study material for students. These solutions are extremely helpful for students who are preparing for competitive exams. The solution is available in PDF format and students can freely download it from our website.

Download PDF of R S Aggarwal Solutions for Class 10 Maths Chapter 13 Trigonometric Identities Exercise 13B

rs aggarwal sol class 10 maths chapter 13
rs aggarwal sol class 10 maths chapter 13
rs aggarwal sol class 10 maths chapter 13
rs aggarwal sol class 10 maths chapter 13
rs aggarwal sol class 10 maths chapter 13
rs aggarwal sol class 10 maths chapter 13
rs aggarwal sol class 10 maths chapter 13

 

Access other exercise solutions of Class 10 Maths Chapter 13 Trigonometric Identities

Exercise 13 A Solutions

Exercise 13 C Solutions

Access solutions to Maths R S Aggarwal Class 10 Chapter 13 – Trigonometric Identities Exercise 13B

Question 1: If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that (m2 + n2) = (a2 + b2).

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: If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x2 − y2) = (a2 − b2).

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:

rs aggarwal class 10 chapter 13b 1

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 cos θ sin θ = 1 ….(1)

x2/a2 cos2 θ + y2/b2 sin2 θ + 2 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: If (sec θ + tan θ) = m and (sec θ − tan θ) = n, show that mn = 1.

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: If (cosec θ + cot θ) = m and (cosec θ − cot θ) = n, show that mn = 1.

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: If x = a cos3 θ and y = b sin3 θ, prove that

rs aggarwal class 10 chapter 13b 2

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: If (tan θ + sin θ) = m and (tan θ − sin θ) = n, prove that (m2 − n2)2 = 16mn.

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: If (cot θ + tan θ) = m and (sec θ − cos θ) = n, prove that (m2n)^(2/3) − (mn2)^(2/3) = 1.

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

prove that a2b2(a2 + b2) = 1.

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: If (2 sin θ + 3 cos θ) = 2, show that (3 sin θ − 2 cos θ) = ± 3.

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.

R S Aggarwal Solutions for Class 10 Maths Chapter 13 Trigonometric Identities Exercise 13B

Class 10 Maths Chapter 13 Trigonometric Identities Exercise 13B is based on Elimination of trigonometric ratios. Download R S Aggarwal Solutions Class 10 Chapter 13 and score well in your examinations.