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.

 

Access Answers to Maths R S Aggarwal Chapter 12 Trigonometric Ratios of Complementary Angles Exercise 12

Exercise 12 Page No: 563

Question 1.

rs aggarwal class 10 chapter 12 img 1

rs aggarwal class 10 chapter 12 img 2

Question 2:

Solution:

(i) LHS = cos81° – sin9°

= cos(90° -9°)- sin9°

= sin9° – sin9°

= 0

= RHS

(ii) LHS = tan71° – cot19°

=tan(90° – 19°) – cot19°

=cot19° – cot19°

=0

= RHS

(iii) LHS = cosec80° – sec10°

= cosec(90° – 10°) – sec(10°)

= sec10° – sec10°

= 0

= RHS

(iv) cosec^2 72° − tan^2 18° = 1

LHS = cosec^2 72° – tan^2 18°

= cosec^2 72° – tan2 (90 – 72)°

= cosec^2 72° – cot^2 72°

= 1 = RHS

(v) cos^2 75° + cos^2 15° = 1

LHS = cos^2 75° + cos^2 15°

= cos^2 75° + cos^2 (90 – 75)°

= cos^2 75° + sin^2 75°

= 1= RHS

(vi) tan^2 66° − cot^2 24° = 0

LHS = tan^2 66° − cot^2 24°

= tan^2 66° – cot^2 (90 – 66)°

= tan^2 66° – tan^2 66°

= 0

= RHS

(vii) sin^2 48° + sin^2 42° = 1

LHS = sin^2 48° + sin^2 42°

= sin^2 48° + sin^2 (90 – 48)°

= sin^2 48° + cos^2 48°

= 1

= RHS

(viii) cos^2 57° − sin^2 33° = 0

LHS = cos^2 57° − sin^2 33°

= cos^2 57° – sin^2 (90 – 57)°

= cos^2 57° – cos^2 57°

= 0

=RHS

(ix) (sin 65° + cos 25°)(sin 65° − cos 25°) = 0

LHS = (sin 65° + cos 25°)(sin 65° − cos 25°)

= sin^2 65° – cos^2 25°

= sin^2 65° – cos^2 (90 – 65)°

= sin^2 65° – sin^2 65°

= 0

=RHS

Question 3.
Solution:

(i) LHS = sin53° cos37° + cos53° sin37°

= sin53° cos(90° – 53°) + cos53° sin (90° – 53°)

= sin53° x sin53° + cos53° x cos53°

= sin^2 53° + cos^2 53°

= 1

= RHS

(ii) LHS = cos54° cos36° − sin54° sin36°

= cos54° cos36° − sin(90° -36°) sin(90° -54°)

= cos54° cos36° – cos36°cos54°

= 0

=RHS

(iii) LHS = sec70° sin20° + cos20° cosec70°

= sec(90° – 20°) sin20° + cos20° cosec(90° – 20°)

= cosec 20° sin20° + cos20° sec 20°

= 1 +1

= 2

=RHS

(iv) LHS = sin35° sin55° − cos35° cos55°

= sin(90° – 55°) sin(90° – 35°) − cos35° cos55°

= cos55° cos35° – cos35° cos55°

= 0

=RHS

(v) LHS = (sin72° + cos18°)(sin72° − cos18°)

=(sin^2 72° – cos^2 18°)

= (sin^2 72° – cos^2 (90° – 72°))

= sin^2 72° – sin^2 72°

= 0

=RHS

(vi) LHS = tan48° tan23° tan42° tan67°

=tan48° tan23° tan (90° – 48°) tan (90° – 23°)

= tan48° tan23° cot48° cot23°

= 1×1

= 1

=RHS

Question 4.

Solution:

rs aggarwal class 10 chapter 12 img 4

rs aggarwal class 10 chapter 12 img 5

(v)

rs aggarwal class 10 chapter 12 img 6

Question 5.
Solution:

(i)

LHS = sin θ cos (90° – θ) + sin (90° – θ) cos θ

= sin θ sin θ + cos θ cos θ

= sin^2 θ + cos^2 θ

= 1

= R.H.S.

Hence proved.

rs aggarwal class 10 chapter 12 img 8

rs aggarwal class 10 chapter 12 img 9

(vi)

rs aggarwal class 10 chapter 12 img 10

=(1+1)/(3×1)

= 2/3 = RHS

Hence proved.

(vii)

LHS = cot θ tan (90° -θ) – sec (90° – θ)cosec θ +√3tan 12°tan 60°tan 78°

= cot θ cot θ – cosec θ cosec θ +√3 tan 60° tan 12° tan 78°

= cot^2 θ – cosec^2 θ +√3 tan 60° tan 12° tan(90-12)°

= – (cosec^2 θ – cot^2 θ) +√3 tan 60° tan 12° cot 12°

= – 1 + √3(√3 × 1)

= – 1 + 3

= 2

= R.H.S.

Hence proved.

Question 6:
Solution:

rs aggarwal class 10 chapter 12 img 11

rs aggarwal class 10 chapter 12 img 12

Question 7:
Solution:

L.H.S.

= sin (70°+θ) — cos (20° — θ)

= sin (70°+θ) — cos [90°-(70° + θ)]

= sin (70°+θ) — sin (70° + θ)

= 0

= R.H.S.

Hence Proved.

(ii)

L.H.S.

= tan (55° — θ) — cot (35° + θ)

= tan (90°-(35° +θ)) — cot (35° + θ)

= cot (35° + θ) – cot (35° + θ)

= 0

=RHS

Hence Proved.

(iii)

L.H.S.

= cosec (67° + θ) — sec (23° — θ)

= cosec (67° + θ) – sec (90° —(23° + θ))

= cosec (67° + θ) – cosec (67° + θ)

= 0

=RHS

Hence Proved.

(iv)

L.H.S.

= cosec (65° + θ) — sec (25° — θ) — tan (55° — θ) + cot (35° + θ)

= cosec (65° + θ) — sec (90° – (65° + θ)) — tan (90° – (35° + θ)) + cot (35° + θ)

= cosec (65° + θ) — cosec (65° + θ)) — cot (35° + θ) + cot (35° + θ)

= 0

= R.H.S.

(v)

L.H.S.

= sin (50° +θ) — cos (40° — θ) + tan 1° tan 10° tan 80° tan 89°

= sin ((90°-(40° – θ)) — cos (40° — θ) + (tan 1° tan 89°)(tan10° tan 80°)

= cos (40° – θ) – cos (40° – θ) + {tan 1° tan (90°-1°)}{tan10° tan(90°-10°)}

= 0 + {(tan 1° cot 1°}{tan10° cot 10°}

= 0 + 1 = 1

= R.H.S.

Question 8.

Solution:

rs aggarwal class 10 chapter 12 img 13

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 – B

And, (A + C)/2 = (180o – B)/2 = 90o – B/2

Now, tan (A + C)/2 = tan(90o – B/2) = cot B/2

Hence Proved.

Question 10.
Solution:

cos 2θ = sin 4θ …(1)

We know that,

sin( 90° – θ) = cos θ

So, equation (1) can be written as

sin (90° – 2θ) = sin 4θ

On comparing both sides

90° – 2θ = 4θ

90° = 4θ + 2θ

6θ = 90°

or θ = 15°

The value of θ is 15°

Question 11:
Solution: Given: sec2A = cosec(A − 42°)

rs aggarwal class 10 chapter 12 img 14

The value of angle A is 44 degrees.

Question 12:
Solution:

sin 3 A = cos (A − 26°) (given)

or cos (90° – 3A) = cos (A – 26°)

On comparing

90° – 3A = A – 26°

A + 3A = 90° + 26°

4A = 116° = 29°

The value of A is 29°.

Question 13:
Solution:

tan 2A = cot (A – 12°)

or cot (90° – 2A) = cot (A – 12°)

On comparing

90° – 2A = A – 12 °

A + 2A = 90° + 12°

3A = 102°

A = 34°

The value of A is 34°

Question 14:
Solution: sec 4A = cosec (A – 15°)

or cosec (90° – 4A) = cosec (A – 15°)

On comparing

90° – 4A = A – 15°

A + 4A = 90° + 15°

5A = 105°

A = 21°

The value of A is 21°.

Question 15:

Solution:

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

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

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

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

= 2/3 [cosec^2 58°- cot^2 58°] – 5/3

= (2/3) – (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.

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