# R S Aggarwal Solutions for Class 10 Maths Chapter 12 Trigonometric Ratios of Complementary Angles

R S Aggarwal Class 10 Solutions for Chapter 12 Trigonometric ratios of Complementary Angles, detailed explanations to the questions are available here. This chapter is composed of only one exercise which helps students to solve the trigonometric ratios without using trigonometric tables. Click on the link below to get your R S Aggarwal solutions pdf for free and build your skills.

## Exercise 12 Page No: 563

Question 1.

Question 2:

Solution:

(i) LHS = cos81Â° â€“ sin9Â°

= cos(90Â° -9Â°)- sin9Â°

= sin9Â° â€“ sin9Â°

= 0

= RHS

(ii) LHS = tan71Â° â€“ cot19Â°

=tan(90Â° â€“ 19Â°) â€“ cot19Â°

=cot19Â° â€“ cot19Â°

=0

= RHS

(iii) LHS = cosec80Â° â€“ sec10Â°

= cosec(90Â° â€“ 10Â°) â€“ sec(10Â°)

= sec10Â° â€“ sec10Â°

= 0

= RHS

(iv) cosec^2 72Â° âˆ’ tan^2 18Â° = 1

LHS = cosec^2 72Â° – tan^2 18Â°

= cosec^2 72Â° – tan2 (90 – 72)Â°

= cosec^2 72Â° – cot^2 72Â°

= 1 = RHS

(v) cos^2 75Â° + cos^2 15Â° = 1

LHS = cos^2 75Â° + cos^2 15Â°

= cos^2 75Â° + cos^2 (90 – 75)Â°

= cos^2 75Â° + sin^2 75Â°

= 1= RHS

(vi) tan^2 66Â° âˆ’ cot^2 24Â° = 0

LHS = tan^2 66Â° âˆ’ cot^2 24Â°

= tan^2 66Â° – cot^2 (90 – 66)Â°

= tan^2 66Â° – tan^2 66Â°

= 0

= RHS

(vii) sin^2 48Â° + sin^2 42Â° = 1

LHS = sin^2 48Â° + sin^2 42Â°

= sin^2 48Â° + sin^2 (90 – 48)Â°

= sin^2 48Â° + cos^2 48Â°

= 1

= RHS

(viii) cos^2 57Â° âˆ’ sin^2 33Â° = 0

LHS = cos^2 57Â° âˆ’ sin^2 33Â°

= cos^2 57Â° – sin^2 (90 – 57)Â°

= cos^2 57Â° – cos^2 57Â°

= 0

=RHS

(ix) (sin 65Â° + cos 25Â°)(sin 65Â° âˆ’ cos 25Â°) = 0

LHS = (sin 65Â° + cos 25Â°)(sin 65Â° âˆ’ cos 25Â°)

= sin^2 65Â° – cos^2 25Â°

= sin^2 65Â° – cos^2 (90 – 65)Â°

= sin^2 65Â° – sin^2 65Â°

= 0

=RHS

Question 3.
Solution:

(i) LHS = sin53Â° cos37Â° + cos53Â° sin37Â°

= sin53Â° cos(90Â° – 53Â°) + cos53Â° sin (90Â° – 53Â°)

= sin53Â° x sin53Â° + cos53Â° x cos53Â°

= sin^2 53Â° + cos^2 53Â°

= 1

= RHS

(ii) LHS = cos54Â° cos36Â° âˆ’ sin54Â° sin36Â°

= cos54Â° cos36Â° âˆ’ sin(90Â° -36Â°) sin(90Â° -54Â°)

= cos54Â° cos36Â° – cos36Â°cos54Â°

= 0

=RHS

(iii) LHS = sec70Â° sin20Â° + cos20Â° cosec70Â°

= sec(90Â° – 20Â°) sin20Â° + cos20Â° cosec(90Â° – 20Â°)

= cosec 20Â° sin20Â° + cos20Â° sec 20Â°

= 1 +1

= 2

=RHS

(iv) LHS = sin35Â° sin55Â° âˆ’ cos35Â° cos55Â°

= sin(90Â° – 55Â°) sin(90Â° – 35Â°) âˆ’ cos35Â° cos55Â°

= cos55Â° cos35Â° – cos35Â° cos55Â°

= 0

=RHS

(v) LHS = (sin72Â° + cos18Â°)(sin72Â° âˆ’ cos18Â°)

=(sin^2 72Â° – cos^2 18Â°)

= (sin^2 72Â° – cos^2 (90Â° – 72Â°))

= sin^2 72Â° – sin^2 72Â°

= 0

=RHS

(vi) LHS = tan48Â° tan23Â° tan42Â° tan67Â°

=tan48Â° tan23Â° tan (90Â° – 48Â°) tan (90Â° – 23Â°)

= tan48Â° tan23Â° cot48Â° cot23Â°

= 1×1

= 1

=RHS

Question 4.

Solution:

(v)

Question 5.
Solution:

(i)

LHS = sin Î¸ cos (90Â° – Î¸) + sin (90Â° – Î¸) cos Î¸

= sin Î¸ sin Î¸ + cos Î¸ cos Î¸

= sin^2 Î¸ + cos^2 Î¸

= 1

= R.H.S.

Hence proved.

(vi)

=(1+1)/(3×1)

= 2/3 = RHS

Hence proved.

(vii)

LHS = cot Î¸ tan (90Â° -Î¸) – sec (90Â° – Î¸)cosec Î¸ +âˆš3tan 12Â°tan 60Â°tan 78Â°

= cot Î¸ cot Î¸ – cosec Î¸ cosec Î¸ +âˆš3 tan 60Â° tan 12Â° tan 78Â°

= cot^2 Î¸ â€“ cosec^2 Î¸ +âˆš3 tan 60Â° tan 12Â° tan(90-12)Â°

= – (cosec^2 Î¸ â€“ cot^2 Î¸) +âˆš3 tan 60Â° tan 12Â° cot 12Â°

= – 1 + âˆš3(âˆš3 Ã— 1)

= – 1 + 3

= 2

= R.H.S.

Hence proved.

Question 6:
Solution:

Question 7:
Solution:

L.H.S.

= sin (70Â°+Î¸) â€” cos (20Â° â€” Î¸)

= sin (70Â°+Î¸) â€” cos [90Â°-(70Â° + Î¸)]

= sin (70Â°+Î¸) â€” sin (70Â° + Î¸)

= 0

= R.H.S.

Hence Proved.

(ii)

L.H.S.

= tan (55Â° â€” Î¸) â€” cot (35Â° + Î¸)

= tan (90Â°-(35Â° +Î¸)) â€” cot (35Â° + Î¸)

= cot (35Â° + Î¸) – cot (35Â° + Î¸)

= 0

=RHS

Hence Proved.

(iii)

L.H.S.

= cosec (67Â° + Î¸) â€” sec (23Â° â€” Î¸)

= cosec (67Â° + Î¸) – sec (90Â° â€”(23Â° + Î¸))

= cosec (67Â° + Î¸) – cosec (67Â° + Î¸)

= 0

=RHS

Hence Proved.

(iv)

L.H.S.

= cosec (65Â° + Î¸) â€” sec (25Â° â€” Î¸) â€” tan (55Â° â€” Î¸) + cot (35Â° + Î¸)

= cosec (65Â° + Î¸) â€” sec (90Â° – (65Â° + Î¸)) â€” tan (90Â° – (35Â° + Î¸)) + cot (35Â° + Î¸)

= cosec (65Â° + Î¸) â€” cosec (65Â° + Î¸)) â€” cot (35Â° + Î¸) + cot (35Â° + Î¸)

= 0

= R.H.S.

(v)

L.H.S.

= sin (50Â° +Î¸) â€” cos (40Â° â€” Î¸) + tan 1Â° tan 10Â° tan 80Â° tan 89Â°

= sin ((90Â°-(40Â° – Î¸)) â€” cos (40Â° â€” Î¸) + (tan 1Â° tan 89Â°)(tan10Â° tan 80Â°)

= cos (40Â° – Î¸) – cos (40Â° – Î¸) + {tan 1Â° tan (90Â°-1Â°)}{tan10Â° tan(90Â°-10Â°)}

= 0 + {(tan 1Â° cot 1Â°}{tan10Â° cot 10Â°}

= 0 + 1 = 1

= R.H.S.

Question 8.

Solution:

Question 9.
Solution:

Given function is : $tan\frac{C+A}{2} = cot\frac{B}{2}$

Sum of all the angles of a triangle = 180 degree

So, A + B + C = 180o

Or A + C = 180o â€“ B

And, (A + C)/2 = (180o â€“ B)/2 = 90o â€“ B/2

Now, tan (A + C)/2 = tan(90o â€“ B/2) = cot B/2

Hence Proved.

Question 10.
Solution:

cos 2Î¸ = sin 4Î¸ …(1)

We know that,

sin( 90Â° – Î¸) = cos Î¸

So, equation (1) can be written as

sin (90Â° – 2Î¸) = sin 4Î¸

On comparing both sides

90Â° – 2Î¸ = 4Î¸

90Â° = 4Î¸ + 2Î¸

6Î¸ = 90Â°

or Î¸ = 15Â°

The value of Î¸ is 15Â°

Question 11:
Solution: Given: sec2A = cosec(A âˆ’ 42Â°)

The value of angle A is 44 degrees.

Question 12:
Solution:

sin 3 A = cos (A âˆ’ 26Â°) (given)

or cos (90Â° – 3A) = cos (A – 26Â°)

On comparing

90Â° – 3A = A – 26Â°

A + 3A = 90Â° + 26Â°

4A = 116Â° = 29Â°

The value of A is 29Â°.

Question 13:
Solution:

tan 2A = cot (A â€“ 12Â°)

or cot (90Â° – 2A) = cot (A â€“ 12Â°)

On comparing

90Â° – 2A = A â€“ 12 Â°

A + 2A = 90Â° + 12Â°

3A = 102Â°

A = 34Â°

The value of A is 34Â°

Question 14:
Solution: sec 4A = cosec (A â€“ 15Â°)

or cosec (90Â° – 4A) = cosec (A â€“ 15Â°)

On comparing

90Â° – 4A = A â€“ 15Â°

A + 4A = 90Â° + 15Â°

5A = 105Â°

A = 21Â°

The value of A is 21Â°.

Question 15:

Solution:

= 2/3 cosec^2 58Â°- 2/3 cot 58Â° tan 32Â° – 5/3 tan 13Â° tan 37Â° tan 45Â° tan 53Â° tan 77Â°

= 2/3 cosec^2 58Â°- 2/3 cot 58Â° tan (90-58)Â° – 5/3 tan 45Â° (tan 13Â° tan 77Â°) (tan 37Â° tan 53Â°)

= 2/3 cosec^2 58Â°- 2/3 cot 58Â° cot 58Â° – 5/3 Ã— 1 x (tan 13Â° tan (90-13)Â°) Ã— (tan 37Â° tan (90-37)Â°)

= 2/3 cosec^2 58Â°- 2/3 cot^2 58Â° – 5/3 Ã— (tan 13Â° cot 13Â°) (tan 37Â° cot 37Â°)

= 2/3 [cosec^2 58Â°- cot^2 58Â°] – 5/3

= (2/3) â€“ (5/3)

= -1

= R.H.S.

## R S Aggarwal Solutions for Chapter 12 Trigonometric Ratios of Complementary Angles Topics

In this chapter students will study important concepts on the topic as listed below:

• Complementary angles introduction.

Two angles are complementary angles if their sum is 90 degrees. For example, if one angle is x^o then second angle will be (90-x)^o.

• Learn to evaluate various trigonometric ratios without using trigonometric tables.

RS Aggarwal Solutions for Class 10 Chapter 12 Trigonometric ratios of complementary angles provides step by step answers, which help students understand the concepts easily. Download and practice R S Aggarwal Solutions for Maths to score more in your exams.

### Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 12 Trigonometric ratios of Complementary Angles

1. R S Aggarwal provides conceptual knowledge on the topic trigonometric ratios of complementary angles.

2. Helps to crack competitive exams.

3. Set of various problems along with their answers.

4. All questions are answered step by step by subject experts.