**NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry**

**NCERT Solutions** of **important questions** from **Class 10 Maths**** Chapter 8 ****Introduction To Trigonometry **are given here straightforward and well-ordered clarifications. These answers for **Introduction To** **Trigonometry** are amazingly well known among **Class 10** students for **Math Introduction To Trigonometry Solutions** convenient for rapidly finishing your homework and getting ready for tests. All questions and answers from the **NCERT **Book of **Class 10 Maths** **Chapter 8 **are given here to you . You will likewise adore the advertisement-free understanding on **future study point’s NCERT Solutions** of **Important** **questions.** All **NCERT Solutions of Important Questions for Class 10 Maths** are set up by specialists and are **100% precise**.

Click for online shopping

**Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc**

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

**Exercise 8.1- Introduction to Trigonometry**

**Exercise 8.2 -Introduction to Trigonometry**

**Exercise 8.3-Introduction to Trigonometry**

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

**NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry **

For doing all the trigonometric solutions you should have the following formulas in your brain**.**

Point to be noticed that here perpendicular means side opposite to the angle and base means side adjacent to the angle, the hypotenuse is the longest side opposite to right angle as in the following fig.

, ,

**NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry **

**Example –**

Express cosθ into all other trigonometric ratios.

Solution-

transformation into sinθ

The relationship between cosθ and sinθ is given as

sin²θ + cos²θ = 1

cos²θ = 1 – sin²θ

Transformation into tanθ

There is no direct relation between cosθ and tanθ but we know the relationship between cosθ and secθ as following.

…………(1)

Determining the value of secθ in terms of tanθ.

1 + tan²θ = sec²θ

………(2)

Substituting the value of secθ in (1)

Transformation into cotθ

We have got the cosθ in terms of tanθ

………(1)

We know

………..(2)

Substituting this value of tanθ in (1)

Transformation into secθ

There is a direct relationship between cosθ and secθ

Transformation into cosecθ

While expressing cosθ into cosecθ, think over the following fact

………..(1)

Substituting the value of sinθ in (1)

**NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry **

**Practice exercise-**

(1) express sinθ in terms of tanθ.

(2) Write tan θ in phrases of cosecθ.

(3) decide the relationship between secθ and sinθ.

(4) remodel cosecθ into all trigonometric ratios.

(5)express cotθ in terms of secθ.

**(1) Hint:** sinθ⇒cosecθ⇒cotθ⇒tanθ

**(2)Hint**:tanθ⇒cotθ⇒cosecθ

**(3)Hint:**secθ⇒cosθ⇒sinθ

**(4)Hint:** Check the relationship between cosecθ to other trigonometric ratios one by one.

**(5) Hint**:cotθ⇒tanθ⇒secθ

#### Trigonometric Ratios of Complementary Angles-

∠A + ∠B + ∠C = 18O°

∠A + 90° + ∠C =180°

∠A + ∠C = 180°– 90°

∠A + ∠C = 90°

∠A & ∠C are complementry to each other

∠C = 90 –∠A

So, comparing the trigonometric ratios corresponding to ∠A, and ∠(90 –A)i.e∠C.

……………(1)

……(2)

From the equations (1) and (2) we have

sin(90–A) = cosA, cos (90– A) = sinA, tan(90–A) = cotA

cot(90–A) = tanA, sec (90– A) = cosecA, cosec(90–A) = sinA

Determining the value of other trigonometric ratio when value of one of them is given-

**Table of trigonometric functions for cbse 10 th class**

**Study NCERT solutions for class 10 maths exercise 8.1 Introduction to trigonometry**

Example-

Find the value of tanθ if

.

sinθ = 7 : 25

Let perpendicular= 7x and hypotenuse= 25x

Ans.

**NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry **

**Practice exercise- Evaluate the following.**

**(3)**** In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.**

**Study A to Z NCERT 10 th class maths solution of lesson 9.1 ‘Some Application Of the Trigonometry’**

**67° + 23° = 90° and 55° +25° = 90°⇒23= 90 –67 and 25 =90–55, so substituting th values of 23 and 25 in the given expression**

**(sin²θ +cos²θ=1)**

**(2) If ∠A and ∠B are acute angles such that cosA = cosB then show that ∠A = ∠B.**

Answer- Let ΔABC is a right triangle in which ∠A and ∠B are acute angles.

We are given that cosA = cosB

**AC = BC**

∠B = ∠A(angles to equal opposite sides)

See the video for solutions of Q5 part(i),part(ii) and part(iii) of exercise 8.4 class 10 NCERT maths

**L.H.S**

Substituting sin²θ = 1 – cos²θ

= R.H.S

Solution-

L.H.S

= **2cosecθ= R.H.S**

(**5) Prove that ; ****(sinα + cosα)(tanα + cotα)**** = (secα + cosecα)**

Solution-

L.H.S

(sinα + cosα)(tanα + cotα)

= cosecα + secα= R.H.S

**(6) Prove that**

Solution-

L.H.S

= R.H.S

**(7) Prove that**

Since in the R.H.S we have to get secθ and tanθ so transforming L.H.S in terms of secθ and tanθ, dividing the numerator and denominator by cosθ, we get

Let the [tanθ – secθ] remain in its place at the denominator since in the R.H.S we are to be required this one.

So, multiplying the expressions in the numerators

= R.H.S

**(8) Prove that:**

L.H.S

= 2secA = R.H.S

**(9)**

Solution-

L.H.S

=1 + secθ.cosecθ= R.H.S

**(10) Prove that:**

**L.H.S**

= secA + tanA = R.H.S

**(11) Prove that:**

L.H.S

**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions for class 9 maths**

**NCERT Solutions for class 9 science **

**NCERT Solutions for class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**NCERT Solutions for Class 10 Science**

**NCERT Solutions for class 11 maths**

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 | Chapter 16- Probability |

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

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions for class 12 maths**

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 |

**Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2**

**Class 12 Maths Important Questions-Application of Integrals**

**Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22**

**Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution**