NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 8 **Introduction to Trigonometry Class 10 NCERT Solutions Ex 8.4.**

- Introduction to Trigonometry Class 10 Ex 8.1
- Introduction to Trigonometry Class 10 Ex 8.2
- Introduction to Trigonometry Class 10 Ex 8.3

Board | CBSE |

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 8 |

Chapter Name | Introduction to Trigonometry |

Exercise | Ex 8.4 |

Number of Questions Solved | 5 |

Category | NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

NCERT Solutions for Class 10 Maths

**Page No : 193**

**Question 1.** Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

**Solution: **

**Question 2.** Write all the other trigonometric ratios of ∠A in terms of sec A.

**Solution: **

**Question 3.** Evaluate :

(i) \(\frac { { sin }^{ 2 }{ 63 }^{ 0 }+{ sin }^{ 2 }{ 27 }^{ 0 } }{ { cos }^{ 2 }{ 17 }^{ 0 }+{ cos }^{ 2 }{ 73 }^{ 0 } } \)

(ii) sin 25° cos 65° + cos 25° sin 65°

**Solution: **

**Question 4.** Choose the correct option. Justify your choice.

(i) 9 sec^{2}A – 9 tan^{2}A =

(A) 1 (B) 9 (C) 8 (D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

(A) 0 (B) 1 (C) 2 (D) – 1

(iii) (secA + tanA) (1 – sinA) =

(A) secA (B) sinA (C) cosecA (D) cosA

(iv) \(\frac { 1+{ tan }^{ 2 }A }{ 1+{ cot }^{ 2 }A } =\)

(A) sec^{2}A (B) -1 (C) cot^{2}A (D) tan^{2}A

**Solution: **(i) 9sec2A – 9tan2A

= 9(sec2A – tan2A)

= 9 (1) [as sec2 A – tan2 A = 1]

= 9

Hence alternative (B) is correct.

(ii) (1 + tanθ + secθ) (1 + cotθ – cosecθ)

Hence alternative (C) is correct.

(iii) (secA + tanA) (1 – sinA)

(iv) \(\frac { 1+{ tan }^{ 2 }A }{ 1+{ cot }^{ 2 }A } =\)

**Question 5.** Prove the following identities, where the angles involved are acute angles for which the

expressions are defined.

(i) \(\left( cosec\theta -cot\theta \right) ^{ 2 }=\frac { 1-cos\theta }{ 1+cos\theta } \)

(ii) \(\frac { cosA }{ 1+sinA } +\frac { 1+sinA }{ cosA } =2secA\)

(iii) \(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } =1+sec\theta cosec\theta \)

(iv) \(\frac { 1-secA }{ secA } =\frac { { sin }^{ 2 }A }{ 1-cosA } \)

(v) \(\frac { cosA-sinA+1 }{ cosA+sinA-1 } =cosecA+cotA\) using the identity cosec^{2}A = 1+cot^{2}A.

(vi) \(\sqrt { \frac { 1+sinA }{ 1-sinA } } =\quad secA+tanA\)

(vii) \(\frac { sin\theta -{ 2sin }^{ 3 }\theta }{ { 2cos }^{ 3 }\theta -cos\theta } =tan\theta \)

(viii) (sin A + cosec A)^{2 }+ (cos A + sec A)^{2} = 7+tan^{2}A+cot^{2}A

(ix) \(\left( cosecA-sinA \right) \left( secA-cosA \right) =\frac { 1 }{ tanA+cotA } \)

(x) \(\left( \frac { 1+{ tan }^{ 2 }A }{ 1+cot^{ 2 }A } \right) ^{ 2 }=\left( \frac { 1-{ tan }A }{ 1-cotA } \right) ^{ 2 }={ tan }^{ 2 }A\)

**Solution:**

(v) \(\frac { cosA-sinA+1 }{ cosA+sinA-1 } =cosecA+cotA\) using the identity cosec^{2}A = 1+cot^{2}A.

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