8.4 class 10

Class 10 Maths NCERT Solutions for Chapter 8 Introduction To Trigonometry Exercise 8.4 provided here is very essential for class 10 students that will helpful in understanding the basic terms and formula to solve problems. It will improve problem solving skills of students as well as boost their confidence level. NCERT Revision Notes for Class 10 will also help in revising the important points of chapter before examinations. These

tan A = 1 / cot A tan A = cot -1 A Question 2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution: (i) cos A cos A = 1/sec A (ii) sin A We know that sin 2A = 1 – cos 2A Also , cos 2A = 1 / sec 2A sin 2A = 1 – 1 / sec 2A sin 2A = (sec 2A – 1) / sec 2A sin A = (sec 2A – 1) 1/2 / sec A (iii) tan A We know that tan 2A + 1 = sec 2A tan A = (sec 2A – 1)½ (iv) cosec A We know cosec A = 1/ sinA cosec A = sec A / (sec 2A – 1)½ (v) cot A We know cot A = cos A / sin A cot A = (1/sec A) / ((sec 2A – 1) 1/2 / sec A) cot A = 1 / (sec 2A – 1) 1/2 Question 3. Evaluate: (i) (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°) (ii) sin 25° cos 65° + cos 25° sin 65° (i) ([sin(90-27)] 2 + sin 2 27) / ([cos(90-73)] 2 + cos 2 73) We know that sin(90-x) = cos x cos(90-x) = sin x (cos 2(27) + sin 2 27) / (sin 2(73) + cos 2 73) Using sin 2A + cos 2A = 1 1/1 = 1 (ii) [sin 25 * cos(90-25)] + [cos 25 * sin(90-25)] Using sin(90-x) = cos x cos(90-x) = sin x = [sin 25 * sin 25] + [cos 25 * cos 25] = sin 2 25 + cos 2 25 = 1 Question 4. Choose the correct option. Justify your choice. Solution: (i) 9 sec 2 A – 9 tan 2 A ( A) 1 (B) 9 (C) 8 (D) 0 Using sec 2A – tan 2A = 1 9 (sec 2A – tan 2A ) = 9(1) Ans (B) (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (A) 0 (B) 1 (C) 2 (D) –1 Simplifying all ratios = (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ) = ((cosθ + sinθ + 1)/ cosθ) ((sinθ + cosθ – 1 )/sinθ) = ((cosθ + sinθ) 2– 1) / (sinθ cosθ) = (1 + 2*cosθ*sinθ – 1) / (sinθ cosθ) = 2 Ans (C)...

NCERT Solutions for Class 10 Maths NCERT solutions for class 10 maths comprise a detailed and step-wise explanation to all NCERT problems for CBSE 10 th grade. Students can use these solutions to help them proceed if they hit a roadblock while attempting a sum. These class 10 maths NCERT solutions are prepared by a host of math experts who are wizards in this field. The solutions are presented in such a way that students not only get the answer to a specific problem but can also develop a clear understanding of the concept used in that question. NCERT solutions for class 10 maths are extremely helpful for students preparing for the 10 th board examinations. The examination paper follows the structure provided by NCERT hence, it becomes very important for students to refer to these solutions. They also build a foundation for topics that will be covered in the upcoming 11th and 12th grades. CBSE solutions for class 10 maths are useful in preparing for other competitive exams such as Olympiads. 1. 2. 3. 4. 5. 6. 7. 8 9. 10. 11. 12. 13. 14. 15. 16. 17. NCERT Solutions for Class 10 Maths Updated for 2022-23 Session – Free PDF Download NCERT solutions for class th boards. An in-depth analysis of the 10 th maths CBSE solutions sums is available for free pdf download. So start your impressive educational journey today! The links to all the chapter-wise pdfs are given below. Class 10 Maths NCERT Solutions Chapter 1 to 15 Download NCERT Solutions of Class 10 Maths Chapter-wise PDF P...

tan A = 1 / cot A tan A = cot -1 A Question 2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution: (i) cos A cos A = 1/sec A (ii) sin A We know that sin 2A = 1 – cos 2A Also , cos 2A = 1 / sec 2A sin 2A = 1 – 1 / sec 2A sin 2A = (sec 2A – 1) / sec 2A sin A = (sec 2A – 1) 1/2 / sec A (iii) tan A We know that tan 2A + 1 = sec 2A tan A = (sec 2A – 1)½ (iv) cosec A We know cosec A = 1/ sinA cosec A = sec A / (sec 2A – 1)½ (v) cot A We know cot A = cos A / sin A cot A = (1/sec A) / ((sec 2A – 1) 1/2 / sec A) cot A = 1 / (sec 2A – 1) 1/2 Question 3. Evaluate: (i) (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°) (ii) sin 25° cos 65° + cos 25° sin 65° (i) ([sin(90-27)] 2 + sin 2 27) / ([cos(90-73)] 2 + cos 2 73) We know that sin(90-x) = cos x cos(90-x) = sin x (cos 2(27) + sin 2 27) / (sin 2(73) + cos 2 73) Using sin 2A + cos 2A = 1 1/1 = 1 (ii) [sin 25 * cos(90-25)] + [cos 25 * sin(90-25)] Using sin(90-x) = cos x cos(90-x) = sin x = [sin 25 * sin 25] + [cos 25 * cos 25] = sin 2 25 + cos 2 25 = 1 Question 4. Choose the correct option. Justify your choice. Solution: (i) 9 sec 2 A – 9 tan 2 A ( A) 1 (B) 9 (C) 8 (D) 0 Using sec 2A – tan 2A = 1 9 (sec 2A – tan 2A ) = 9(1) Ans (B) (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (A) 0 (B) 1 (C) 2 (D) –1 Simplifying all ratios = (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ) = ((cosθ + sinθ + 1)/ cosθ) ((sinθ + cosθ – 1 )/sinθ) = ((cosθ + sinθ) 2– 1) / (sinθ cosθ) = (1 + 2*cosθ*sinθ – 1) / (sinθ cosθ) = 2 Ans (C)...

NCERT Solutions for Class 10 Maths NCERT solutions for class 10 maths comprise a detailed and step-wise explanation to all NCERT problems for CBSE 10 th grade. Students can use these solutions to help them proceed if they hit a roadblock while attempting a sum. These class 10 maths NCERT solutions are prepared by a host of math experts who are wizards in this field. The solutions are presented in such a way that students not only get the answer to a specific problem but can also develop a clear understanding of the concept used in that question. NCERT solutions for class 10 maths are extremely helpful for students preparing for the 10 th board examinations. The examination paper follows the structure provided by NCERT hence, it becomes very important for students to refer to these solutions. They also build a foundation for topics that will be covered in the upcoming 11th and 12th grades. CBSE solutions for class 10 maths are useful in preparing for other competitive exams such as Olympiads. 1. 2. 3. 4. 5. 6. 7. 8 9. 10. 11. 12. 13. 14. 15. 16. 17. NCERT Solutions for Class 10 Maths Updated for 2022-23 Session – Free PDF Download NCERT solutions for class th boards. An in-depth analysis of the 10 th maths CBSE solutions sums is available for free pdf download. So start your impressive educational journey today! The links to all the chapter-wise pdfs are given below. Class 10 Maths NCERT Solutions Chapter 1 to 15 Download NCERT Solutions of Class 10 Maths Chapter-wise PDF P...

Class 10 Maths NCERT Solutions for Chapter 8 Introduction To Trigonometry Exercise 8.4 provided here is very essential for class 10 students that will helpful in understanding the basic terms and formula to solve problems. It will improve problem solving skills of students as well as boost their confidence level. NCERT Revision Notes for Class 10 will also help in revising the important points of chapter before examinations. These

NCERT Solutions for Class 10 Maths NCERT solutions for class 10 maths comprise a detailed and step-wise explanation to all NCERT problems for CBSE 10 th grade. Students can use these solutions to help them proceed if they hit a roadblock while attempting a sum. These class 10 maths NCERT solutions are prepared by a host of math experts who are wizards in this field. The solutions are presented in such a way that students not only get the answer to a specific problem but can also develop a clear understanding of the concept used in that question. NCERT solutions for class 10 maths are extremely helpful for students preparing for the 10 th board examinations. The examination paper follows the structure provided by NCERT hence, it becomes very important for students to refer to these solutions. They also build a foundation for topics that will be covered in the upcoming 11th and 12th grades. CBSE solutions for class 10 maths are useful in preparing for other competitive exams such as Olympiads. 1. 2. 3. 4. 5. 6. 7. 8 9. 10. 11. 12. 13. 14. 15. 16. 17. NCERT Solutions for Class 10 Maths Updated for 2022-23 Session – Free PDF Download NCERT solutions for class th boards. An in-depth analysis of the 10 th maths CBSE solutions sums is available for free pdf download. So start your impressive educational journey today! The links to all the chapter-wise pdfs are given below. Class 10 Maths NCERT Solutions Chapter 1 to 15 Download NCERT Solutions of Class 10 Maths Chapter-wise PDF P...

Class 10 Maths NCERT Solutions for Chapter 8 Introduction To Trigonometry Exercise 8.4 provided here is very essential for class 10 students that will helpful in understanding the basic terms and formula to solve problems. It will improve problem solving skills of students as well as boost their confidence level. NCERT Revision Notes for Class 10 will also help in revising the important points of chapter before examinations. These

tan A = 1 / cot A tan A = cot -1 A Question 2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution: (i) cos A cos A = 1/sec A (ii) sin A We know that sin 2A = 1 – cos 2A Also , cos 2A = 1 / sec 2A sin 2A = 1 – 1 / sec 2A sin 2A = (sec 2A – 1) / sec 2A sin A = (sec 2A – 1) 1/2 / sec A (iii) tan A We know that tan 2A + 1 = sec 2A tan A = (sec 2A – 1)½ (iv) cosec A We know cosec A = 1/ sinA cosec A = sec A / (sec 2A – 1)½ (v) cot A We know cot A = cos A / sin A cot A = (1/sec A) / ((sec 2A – 1) 1/2 / sec A) cot A = 1 / (sec 2A – 1) 1/2 Question 3. Evaluate: (i) (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°) (ii) sin 25° cos 65° + cos 25° sin 65° (i) ([sin(90-27)] 2 + sin 2 27) / ([cos(90-73)] 2 + cos 2 73) We know that sin(90-x) = cos x cos(90-x) = sin x (cos 2(27) + sin 2 27) / (sin 2(73) + cos 2 73) Using sin 2A + cos 2A = 1 1/1 = 1 (ii) [sin 25 * cos(90-25)] + [cos 25 * sin(90-25)] Using sin(90-x) = cos x cos(90-x) = sin x = [sin 25 * sin 25] + [cos 25 * cos 25] = sin 2 25 + cos 2 25 = 1 Question 4. Choose the correct option. Justify your choice. Solution: (i) 9 sec 2 A – 9 tan 2 A ( A) 1 (B) 9 (C) 8 (D) 0 Using sec 2A – tan 2A = 1 9 (sec 2A – tan 2A ) = 9(1) Ans (B) (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (A) 0 (B) 1 (C) 2 (D) –1 Simplifying all ratios = (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ) = ((cosθ + sinθ + 1)/ cosθ) ((sinθ + cosθ – 1 )/sinθ) = ((cosθ + sinθ) 2– 1) / (sinθ cosθ) = (1 + 2*cosθ*sinθ – 1) / (sinθ cosθ) = 2 Ans (C)...

tan A = 1 / cot A tan A = cot -1 A Question 2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution: (i) cos A cos A = 1/sec A (ii) sin A We know that sin 2A = 1 – cos 2A Also , cos 2A = 1 / sec 2A sin 2A = 1 – 1 / sec 2A sin 2A = (sec 2A – 1) / sec 2A sin A = (sec 2A – 1) 1/2 / sec A (iii) tan A We know that tan 2A + 1 = sec 2A tan A = (sec 2A – 1)½ (iv) cosec A We know cosec A = 1/ sinA cosec A = sec A / (sec 2A – 1)½ (v) cot A We know cot A = cos A / sin A cot A = (1/sec A) / ((sec 2A – 1) 1/2 / sec A) cot A = 1 / (sec 2A – 1) 1/2 Question 3. Evaluate: (i) (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°) (ii) sin 25° cos 65° + cos 25° sin 65° (i) ([sin(90-27)] 2 + sin 2 27) / ([cos(90-73)] 2 + cos 2 73) We know that sin(90-x) = cos x cos(90-x) = sin x (cos 2(27) + sin 2 27) / (sin 2(73) + cos 2 73) Using sin 2A + cos 2A = 1 1/1 = 1 (ii) [sin 25 * cos(90-25)] + [cos 25 * sin(90-25)] Using sin(90-x) = cos x cos(90-x) = sin x = [sin 25 * sin 25] + [cos 25 * cos 25] = sin 2 25 + cos 2 25 = 1 Question 4. Choose the correct option. Justify your choice. Solution: (i) 9 sec 2 A – 9 tan 2 A ( A) 1 (B) 9 (C) 8 (D) 0 Using sec 2A – tan 2A = 1 9 (sec 2A – tan 2A ) = 9(1) Ans (B) (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (A) 0 (B) 1 (C) 2 (D) –1 Simplifying all ratios = (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ) = ((cosθ + sinθ + 1)/ cosθ) ((sinθ + cosθ – 1 )/sinθ) = ((cosθ + sinθ) 2– 1) / (sinθ cosθ) = (1 + 2*cosθ*sinθ – 1) / (sinθ cosθ) = 2 Ans (C)...