**NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry**

NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry are the best study inputs for the preparations of trigonometry questions in the CBSE board exam of 10 class since maximum questions are asked from this exercise 8.4 -Introduction to Trigonometry. All questions of Exercise 8.4 are solved by an expert of maths as per the CBSE norms.ย NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry are explained here beautifully so that every students of 10 standards could understand properly. You can study here science and maths notes, e-books of science and maths and notes for the carrier, preparation of competitive entrance exams.

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**NCERT Solutions for Class 10 Mathsย Chapter 8 Introduction to Trigonometry**

**Exercise 8.1- Introduction to Trigonometry**

**Exercise 8.2 -Introduction to Trigonometry**

**Exercise 8.3-Introduction to Trigonometry**

**Exercise 8.4-Introduction to Trigonometry**

**NCERT solutions of Important Questions-Introduction to Trigonometry**

**Latest Sample paper Class 10 maths for Term 1 2021 CBSE board**

**NCERT Solutions for Class 10 Maths Exercise 8.4 Chapter 8 Introduction to Trigonometry**

**Q1.Express the trigonometric ratios sin A, sec A and tan A in terms of cot A**

Ans.sin A, sec A and tan A can be transformed into the terms of cot A as follows.

Transforming the sin A into cosec A

sin A = 1/cosec A

Applying the identity

cosecยฒ A =ย 1 + cotยฒ A

cosec A = โ( 1 + cotยฒ A)

Substituting the value of cosec A

sec A = โ( 1 + tanยฒA)

tan A = 1/cot A

**Q2.Write all the other trigonometric ratios of โ A in terms of sec A.**

Ans.

sin A = โ( 1 – cosยฒA)

cos A =1/sec A

tan A = โ(secยฒA -1)

cot A = 1/tan A

cosec A = 1/sin A

Q3. Evaluate

(ii) sin 25ยฐcos 65ยฐ+ cos 25ยฐsin 65ยฐ

Ans.

Since sin 63ยฐ = cos (90ยฐ – 63ยฐ) and cos 17ยฐ = sin(90ยฐ – 17ยฐ)

Substituting the value of sin63ยฐ and cos 17ยฐ

= 1/1 = 1

(ii) sin 25ยฐcos 65ยฐ+ cos 25ยฐsin 65ยฐ

Since sin 25ยฐ= cos( 90ยฐ – 25ยฐ) and cos 25ยฐ = sin(90ยฐ – 25ยฐ)

Substituting the value of sin 25ยฐ and cos 25ยฐ

cos( 90ยฐ – 25ยฐ)cos 65ยฐ+ sin(90ยฐ – 25ยฐ)sin 65ยฐ

cos 65ยฐ.cos 65ยฐ +sin 65ยฐ.sin 65ยฐ

cosยฒ 65ยฐ + sinยฒ 65ยฐ = 1

**Q4.Choose the correct option. Justify your choice.**

**(i) 9 secยฒ A โ 9 tanยฒ A = โฆโฆ**

**(A) 1**

**(B) 9**

**(C) 8**

**(D) 0**

Ans. (i) 9 secยฒ A โ 9 tanยฒ A

=9(1 + tanยฒA) – 9 tanยฒ A

= 9 + 9 tanยฒA – 9 tanยฒ A

=9

Hence the answer is (B) 9

**(ii) (1 + tan ฮธ + sec ฮธ) (1 + cot ฮธ โ cosec ฮธ)**

**(A) 0**

**(B) 1**

**(C) 2**

**(D) -1**

Ans.(ii) (1 + tan ฮธ + sec ฮธ) (1 + cot ฮธ โ cosec ฮธ)

Coverting all trigonometric function in tems of sinฮธ and cosฮธ

( 1 + sinฮธ/cos ฮธย + 1/cosฮธ)( 1 + cosฮธ/sinฮธย – 1/sinฮธ)

(cos ฮธ + sinฮธ + 1)/cosฮธ(sinฮธ + cosฮธ – 1)/sinฮธ

[(cos ฮธ + sinฮธ)ยฒ – 1ยฒ]/cos ฮธ sinฮธ

[cosยฒฮธ + sinยฒฮธ+ 2cos ฮธ sinฮธ-1]/cos ฮธ sinฮธ

[1 + 2cos ฮธ sinฮธ-1]/cos ฮธ sinฮธ = 2cos ฮธ sinฮธ/cos ฮธ sinฮธ =2

Hence the answer is

(C) 2

(A) secยฒA

(B)-1

(C)cotยฒA

(D)tanยฒA

Ans.

Applying the identies 1 + tanยฒA = secยฒA and 1 + cotยฒA = cosecยฒA

Hence the answer is (D) tanยฒA

**Q5.Prove the following identities, where the angles involved are acute angles for which the expressions are defined.**

(i) (cosecฮธ – cotฮธ)ยฒ =(1-cosฮธ)/(1+cosฮธ)

Ans.The given identy is

Taking LHS

Converting the trigonometric function into sin and cos

Taking LHS

Taking LHS

Multiplying the numerator and denominator of second fraction by -1

Applying the identity aยณ – bยณ= (a – b)(aยฒ +bยฒ+ab)

Since sinยฒฮธ + cosยฒฮธ = 1

### See the video -class 10 maths exercise 8.4 question 5 part iv

Ans. Taking LHS

simplifyingย LHS

Multiplying the deniminator and numerator by (1-cosฮธ)

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Ans. Taking LHS

Dividing numerator and denominator by sin A

Rearranging the terms of numerator and denominator

Multiplying the numerator and denominator by the conjugate of cotA -cosecA(i.e cotA +cosecA)

Since cotยฒA = cosecยฒA – 1

Ans. Taking the LHS

Multiplying the denominator and numerator by the conjugate of denominator (1 + sinA)

We have to prove

Taking LHS

Taking sinฮธ and cosฮธ from the numerator and denominator respectively

Since sinฮธ/cosฮธ = tanฮธ and sinยฒฮธ = 1 – cosยฒฮธ

Ans. Taking LHS

Applying the identity (a + b)ยฒ = aยฒ + bยฒ + 2ab

=sinยฒA + cosecยฒA + 2sinA.cosecA +cosยฒA + secยฒA + 2cosA.secA

Arranging the terms and applying the identities 1 + cotยฒA = cosecยฒA and 1+tanยฒA = secยฒA

=sinยฒA + cosยฒA + 1 + cotยฒA + 1+tanยฒA +2sinA.1/sinA +2cosA.1/secA

=1 +1 +1 +2+2 + tanยฒA + cotยฒA

=7 + tanยฒA + cotยฒA=RHS

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Ans.Taking LHS

Transforming the LHS into the terms of sin and cos

Dividing and multiplying it by (tanA + cot A)

Ans.Taking

converting the terms into sin and cos

Taking

Hence it is verified that

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Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

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Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | ย Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

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Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |