#### Chapters

Chapter 2: Polynomials

Chapter 3: Linear Equations in two variables

Chapter 4: Triangles

Chapter 5: Trigonometric Ratios

Chapter 6: T-Ratios of some particular angles

Chapter 7: Trigonometric Ratios of Complementary Angles

Chapter 8: Trigonometric Identities

Chapter 9: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive

Chapter 10: Quadratic Equations

Chapter 11: Arithmetic Progression

Chapter 12: Circles

Chapter 13: Constructions

Chapter 14: Height and Distance

Chapter 15: Probability

Chapter 16: Coordinate Geomentry

Chapter 17: Perimeter and Areas of Plane Figures

Chapter 18: Area of Circle, Sector and Segment

Chapter 19: Volume and Surface Area of Solids

## Chapter 7: Trigonometric Ratios of Complementary Angles

### RS Aggarwal solutions for Secondary School Class 10 Maths Chapter 7 Trigonometric Ratios of Complementary AnglesExercise[Pages 7 - 814]

Without using trigonometric tables, evaluate :

`sin 16^circ/cos 74^circ`

Without using trigonometric tables, evaluate :

`sec 11^circ/("cosec" 79^circ)`

Without using trigonometric tables, evaluate :

`tan 27^circ/cot 63^circ`

Without using trigonometric tables, evaluate :

`cos 35^circ/sin 55^circ`

Without using trigonometric tables, evaluate :

`("cosec" 42^circ)/sec 48^circ`

Without using trigonometric tables, evaluate :

`cot 38^circ/tan 52^circ`

Without using trigonometric tables, prove that:

cos 81° − sin 9° = 0

Without using trigonometric tables, prove that:

tan 71° − cot 19° = 0

Without using trigonometric tables, prove that:

cosec 80° − sec 10° = 0

Without using trigonometric tables, prove that:

cosec^{2}72° − tan^{2}18° = 1

Without using trigonometric tables, prove that:

cos^{2}75° + cos^{2}15° = 1

Without using trigonometric tables, prove that:

tan^{2}66° − cot^{2}24° = 0

Without using trigonometric tables, prove that:

sin^{2}48° + sin^{2}42° = 1

Without using trigonometric tables, prove that:

cos^{2}57° − sin^{2}33° = 0

Without using trigonometric tables, prove that:

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

Without using trigonometric tables, prove that:

sin53° cos37° + cos53° sin37° = 1

Without using trigonometric tables, prove that:

cos54° cos36° − sin54° sin36° = 0

Without using trigonometric tables, prove that:

sec70° sin20° + cos20° cosec70° = 2

Without using trigonometric tables, prove that:

sin35° sin55° − cos35° cos55° = 0

Without using trigonometric tables, prove that:

(sin72° + cos18°)(sin72° − cos18°) = 0

Without using trigonometric tables, prove that:

tan48° tan23° tan42° tan67° = 1

Prove that:

`(sin 70^circ)/(cos 20^circ) + ("cosec" 20^circ)/(sec 70^circ) - 2 cos 70^circ "cosec" 20^circ = 0`

Prove that:

`cos 80^circ/(sin 10^circ) + cos 59^circ "cosec" 31^circ = 2`

Prove that:

`(2 "sin" 68^circ)/(cos 10^circ )- (2 cot 15^circ)/(5 tan 75^circ) = ((3 tan 45^circ t an 20^circ tan 40^circ tan 50^circ tan 70^circ)) /5= 1`

Prove that:

`sin 18^circ/(cos 72^circ )+ sqrt(3)(tan 10^circ tan 30^circ tan 40^circ tan50^circ tan 80^circ) `

Prove that:

sin θ cos (90° - θ ) + sin (90° - θ) cos θ = 1

Prove that:

\[\frac{\sin\theta}{\cos(90° - \theta)} + \frac{\cos\theta}{\sin(90° - \theta)} = 2\]

Prove that:

\[\frac{\sin\theta \cos(90^\circ - \theta)\cos\theta}{\sin(90^\circ- \theta)} + \frac{\cos\theta \sin(90^\circ - \theta)\sin\theta}{\cos(90^\circ - \theta)}\]

Prove that:

\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]

Prove that:

\[\frac{\cos(90^\circ - \theta)}{1 + \sin(90^\circ - \theta)} + \frac{1 + \sin(90^\circ- \theta)}{\cos(90^\circ - \theta)} = 2 cosec\theta\]

Prove that:

\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]

Prove that:

\[cot\theta \tan\left( 90° - \theta \right) - \sec\left( 90° - \theta \right)cosec\theta + \sqrt{3}\tan12° \tan60° \tan78° = 2\]

Prove that :

tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]

Prove that:

cot12° cot38° cot52° cot60° cot78° = \[\frac{1}{\sqrt{3}}\]

Prove that:

cos15° cos35° cosec55° cos60° cosec75° = \[\frac{1}{2}\]

Prove that:

cos1° cos2° cos3° ... cos180° = 0

Prove that:

\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]

Prove that

sin (70° + θ) − cos (20° − θ) = 0

Prove that

tan (55° − θ) − cot (35° + θ) = 0

Prove that

cosec (67° + θ) − sec (23° − θ) = 0

Prove that

cosec (65 °+ θ) sec (25° − θ) − tan (55° − θ) + cot (35° + θ) = 0

Prove that

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

Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

sin67° + cos75°

Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

cot65° + tan49°

Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

sec78° + cosec56°

Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

cosec54° + sin72°

If A, B and C are the angles of a ΔABC, prove that tan `((C + "A")/2) = cot B/2`

If cos 20 = sin 4 θ ,where 2 θ and 4 θ are acute angles, then find the value of θ

If sec2A = cosec(A - 42°), where 2A is an acute angle, then find the value of A.

If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.

If tan 2 A = cot (A − 12°), where 2 A is an acute angle, find the value of A.

If sec 4 A = cosec (A − 15°), where 4 A is an acute angle, find the value of A.

## Chapter 7: Trigonometric Ratios of Complementary Angles

## RS Aggarwal solutions for Secondary School Class 10 Maths chapter 7 - Trigonometric Ratios of Complementary Angles

RS Aggarwal solutions for Secondary School Class 10 Maths chapter 7 (Trigonometric Ratios of Complementary Angles) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Secondary School Class 10 Maths solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. RS Aggarwal textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Secondary School Class 10 Maths chapter 7 Trigonometric Ratios of Complementary Angles are Trigonometry, Trigonometric Ratios, Trigonometric Ratios of Some Special Angles, Trigonometric Ratios of Complementary Angles, Trigonometric Identities, Proof of Existence, Relationships Between the Ratios, Trigonometry, Trigonometric Ratios and Its Reciprocal, Trigonometry, Trigonometric Ratios, Trigonometric Ratios of Some Special Angles, Trigonometric Ratios of Complementary Angles, Trigonometric Identities, Proof of Existence, Relationships Between the Ratios, Trigonometry, Trigonometric Ratios and Its Reciprocal.

Using RS Aggarwal Class 10 solutions Trigonometric Ratios of Complementary Angles exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RS Aggarwal Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 10 prefer RS Aggarwal Textbook Solutions to score more in exam.

Get the free view of chapter 7 Trigonometric Ratios of Complementary Angles Class 10 extra questions for Secondary School Class 10 Maths and can use Shaalaa.com to keep it handy for your exam preparation