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

The trigonometric identities between trigonometric functions are equations that are true for only right-angled triangle. R S Aggarwal Solutions for Class 10 Chapter 13 Trigonometric Identities is an important topic for students studying in Class 10. This chapter mainly deals with important trigonometric identities. In order to solve this chapter’s problems quickly, students are advised to remember all the identities. Students can avail the R S Aggarwal Solutions and download the pdf for free.

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

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
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
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 Solutions to Maths R S Aggarwal Chapter 13 – Trigonometric Identities

Get detailed solutions for all the questions listed under the below exercises:

Exercise 13 A Solutions

Exercise 13 B Solutions

Exercise 13 C Solutions

Exercise 13A

Prove each of the following identities:

Question 1:

(i) (1 – cos2θ) cosec2θ = 1

(ii) (1 + cot2θ) sin2θ = 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:

(i) (sec2θ − 1) cot2θ = 1

(ii) (sec2θ − 1) (cosec2θ − 1) = 1

(iii) (1− cos2θ) sec2θ = tan2θ

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

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

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

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: Prove

rs aggarwal class 10 chapter 13a 2

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: Prove

(i) sec θ (1 − sin θ) (sec θ + tan θ) = 1

(ii) sin θ(1 + tan θ) + cos θ(1 + cot θ) = (sec θ + cosec θ)

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 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.


Exercise 13C

Question 1: Write the value of (1-sin2 θ)sec2 θ.

Solution:

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

= 1

Question 2: Write the value of (1-cos2θ)cosec2θ.

Solution:

(1-cos2θ)cosec2θ

= sin2θ x 1/sin2θ

= 1

Question 3: Write the value of (1+tan2θ)cos2θ.

Solution:

(1+tan2θ)cos2θ = sec2 θ x 1/sec2 θ

= 1

Question 4: Write the value of (1+cot2θ)sin2θ.

Solution:

(1+cot2θ)sin2θ = cose2θ x 1/cose2θ

= 1

Question 5: Write the value of sin2θ + 1/(1+tan2θ)

Solution:

sin2θ + 1/(1+tan2θ)

= sin2θ + 1/(sec2θ)

= sin2θ + cos2θ

= 1

Question 6: Write the value of (cot2θ – 1/sin2θ )

Solution:

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

= (cos2θ – 1)/sin2θ

= -sin2θ/sin2θ

= -1

Question 7: Write the value of sinθ cos(90°-θ)+cosθ sin(90°-θ).

Solution:

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

= sin2θ + cos2θ

= 1

Question 8: Write the value of cosec2(90°-θ) – tan2θ.

Solution:

cosec2(90°-θ) – tan2θ = sec2θ – tan2θ

= 1

Question 9: Write the value of sec2θ(1+sinθ)(1-sinθ).

Solution:

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

= sec2θ x cos2θ

= 1/cos2θ x cos2θ

= 1

Question 10: Write the value of cosec2θ(1+cosθ)(1-cosθ).

Solution:

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

= cosec2θ x sin2θ

= cosec2θ x 1/cosec2θ

= 1

Question 11: Write the value of sin2θ cos2θ(1+tan2θ)(1+cot2θ).

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: Write the value of (1+tan2θ)(1+sinθ)(1-sinθ).

Solution:

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

= sec2θ x cos2θ

= 1/cos2θ x cos2θ

= 1

Question 13: Write the value of 3cot2θ – 3cosec2θ.

Solution:

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

= 3 x -1

= -3

Question 14: Write the value of 4tan2θ – 4/cos2θ

Solution:

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

= (4(sin2θ – 1))/cos2θ

= 4 (-cos2θ) / cos2θ

= -4

Question 15: Write the value of (tan2θ – sec2θ) / (cot2θ – cose2θ)

Solution:

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

= 1

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


R S Aggarwal Solutions for Chapter 13 Trigonometric Identities

In this chapter students will study about Trigonometric Identities concepts as listed below:

  • Trigonometric Identities introduction
  • Important Trigonometric Identities List
  • Elimination of trigonometric ratios

Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 13 Trigonometric Identities

1. Free study materials.

2. Easy for quick revision.

3. These solutions will help the students to strengthen their foundation on Trigonometric Identities.