Trigonometry prove that questions class 10

RD Sharma Class 10 Solutions Chapter 6 Trigonometric Identities RD Sharma Class 10 Solutions Trigonometric Identities Exercise 6.1 Prove the following trigonometric identities : Question 1. (1 – cos 2 A) cosec 2 A = 1 Solution: (1 – cos 2 A) cosec 2 A = 1 L.H.S. = (1 – cos 2 A) cosec 2 A = sin 2 A cosec 2 A (∵ 1 – cos 2 A = sin 2 A) = (sin A cosec A) 2 = (l) 2 = 1 = R.H.S. = [cosec A sin A] 2 = (1) 2= 1 = R.H.S. (∵ sin A cosec A = 1 Question 3. tan 2 θ cos 2 θ = 1- cos 2 θ Solution: Question 4. Solution: Question 5. (sec 2 θ – 1) (cosec 2 θ – 1) = 1 Solution: Question 6. Solution: Question 7. Solution: Question 8. Solution: Question 9. Solution: Question 10. Solution: Question 11. Solution: Question 12. Solution: Question 13. Solution: Question 14. Solution: Question 15. Solution: Question 16. tan 2 θ – sin 2 θ = tan 2 θ sin 2 θ Solution: Question 17. (sec θ + cos θ ) (sec θ – cos θ ) = tan 2 θ + sin 2 θ Solution: Question 18. (cosec θ + sin θ) (cosec θ – sin θ) = cot 2 θ + cos 2 θ Solution: Question 19. sec A (1 – sin A) (sec A + tan A) = 1 (C.B.S.E. 1993) Solution: Question 20. (cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1 Solution: Question 21. (1 + tan 2 θ) (1 – sin θ) (1 + sin θ) = 1 Solution: Question 22. sin 2 A cot 2 A + cos 2 A tan 2 A = 1 (C.B.S.E. 1992C) Solution: Question 23. Solution: Question 24. Solution: Question 25. Solution: Question 26. Solution: Question 27. Solution: Question 28. Solution: Question 29. Solution: Question 30. Solution: Question...

Answer: Answer: OR Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Question 14: Evaluate without using trigonometric tables: Answer: Answer: Question 16: Without using trigonometric tables, evaluate Answer: Question 17: Prove the identity: Answer: Question 18: If , show that Answer: Therefore LHS = RHS. Hence proved. Answer: Answer: Answer: Answer: Answer: Question 24: Without using trigonometric tables evaluate : Answer: Given: Question 25: Without using trigonometric tables evaluate : Answer: Given:

RD Sharma Class 10 Solutions Chapter 6 Trigonometric Identities RD Sharma Class 10 Solutions Trigonometric Identities Exercise 6.1 Prove the following trigonometric identities : Question 1. (1 – cos 2 A) cosec 2 A = 1 Solution: (1 – cos 2 A) cosec 2 A = 1 L.H.S. = (1 – cos 2 A) cosec 2 A = sin 2 A cosec 2 A (∵ 1 – cos 2 A = sin 2 A) = (sin A cosec A) 2 = (l) 2 = 1 = R.H.S. = [cosec A sin A] 2 = (1) 2= 1 = R.H.S. (∵ sin A cosec A = 1 Question 3. tan 2 θ cos 2 θ = 1- cos 2 θ Solution: Question 4. Solution: Question 5. (sec 2 θ – 1) (cosec 2 θ – 1) = 1 Solution: Question 6. Solution: Question 7. Solution: Question 8. Solution: Question 9. Solution: Question 10. Solution: Question 11. Solution: Question 12. Solution: Question 13. Solution: Question 14. Solution: Question 15. Solution: Question 16. tan 2 θ – sin 2 θ = tan 2 θ sin 2 θ Solution: Question 17. (sec θ + cos θ ) (sec θ – cos θ ) = tan 2 θ + sin 2 θ Solution: Question 18. (cosec θ + sin θ) (cosec θ – sin θ) = cot 2 θ + cos 2 θ Solution: Question 19. sec A (1 – sin A) (sec A + tan A) = 1 (C.B.S.E. 1993) Solution: Question 20. (cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1 Solution: Question 21. (1 + tan 2 θ) (1 – sin θ) (1 + sin θ) = 1 Solution: Question 22. sin 2 A cot 2 A + cos 2 A tan 2 A = 1 (C.B.S.E. 1992C) Solution: Question 23. Solution: Question 24. Solution: Question 25. Solution: Question 26. Solution: Question 27. Solution: Question 28. Solution: Question 29. Solution: Question 30. Solution: Question...

Answer: Answer: OR Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Question 14: Evaluate without using trigonometric tables: Answer: Answer: Question 16: Without using trigonometric tables, evaluate Answer: Question 17: Prove the identity: Answer: Question 18: If , show that Answer: Therefore LHS = RHS. Hence proved. Answer: Answer: Answer: Answer: Answer: Question 24: Without using trigonometric tables evaluate : Answer: Given: Question 25: Without using trigonometric tables evaluate : Answer: Given:

RD Sharma Class 10 Solutions Chapter 6 Trigonometric Identities RD Sharma Class 10 Solutions Trigonometric Identities Exercise 6.1 Prove the following trigonometric identities : Question 1. (1 – cos 2 A) cosec 2 A = 1 Solution: (1 – cos 2 A) cosec 2 A = 1 L.H.S. = (1 – cos 2 A) cosec 2 A = sin 2 A cosec 2 A (∵ 1 – cos 2 A = sin 2 A) = (sin A cosec A) 2 = (l) 2 = 1 = R.H.S. = [cosec A sin A] 2 = (1) 2= 1 = R.H.S. (∵ sin A cosec A = 1 Question 3. tan 2 θ cos 2 θ = 1- cos 2 θ Solution: Question 4. Solution: Question 5. (sec 2 θ – 1) (cosec 2 θ – 1) = 1 Solution: Question 6. Solution: Question 7. Solution: Question 8. Solution: Question 9. Solution: Question 10. Solution: Question 11. Solution: Question 12. Solution: Question 13. Solution: Question 14. Solution: Question 15. Solution: Question 16. tan 2 θ – sin 2 θ = tan 2 θ sin 2 θ Solution: Question 17. (sec θ + cos θ ) (sec θ – cos θ ) = tan 2 θ + sin 2 θ Solution: Question 18. (cosec θ + sin θ) (cosec θ – sin θ) = cot 2 θ + cos 2 θ Solution: Question 19. sec A (1 – sin A) (sec A + tan A) = 1 (C.B.S.E. 1993) Solution: Question 20. (cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1 Solution: Question 21. (1 + tan 2 θ) (1 – sin θ) (1 + sin θ) = 1 Solution: Question 22. sin 2 A cot 2 A + cos 2 A tan 2 A = 1 (C.B.S.E. 1992C) Solution: Question 23. Solution: Question 24. Solution: Question 25. Solution: Question 26. Solution: Question 27. Solution: Question 28. Solution: Question 29. Solution: Question 30. Solution: Question...

Answer: Answer: OR Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Question 14: Evaluate without using trigonometric tables: Answer: Answer: Question 16: Without using trigonometric tables, evaluate Answer: Question 17: Prove the identity: Answer: Question 18: If , show that Answer: Therefore LHS = RHS. Hence proved. Answer: Answer: Answer: Answer: Answer: Question 24: Without using trigonometric tables evaluate : Answer: Given: Question 25: Without using trigonometric tables evaluate : Answer: Given:

Answer: Answer: OR Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Question 14: Evaluate without using trigonometric tables: Answer: Answer: Question 16: Without using trigonometric tables, evaluate Answer: Question 17: Prove the identity: Answer: Question 18: If , show that Answer: Therefore LHS = RHS. Hence proved. Answer: Answer: Answer: Answer: Answer: Question 24: Without using trigonometric tables evaluate : Answer: Given: Question 25: Without using trigonometric tables evaluate : Answer: Given:

RD Sharma Class 10 Solutions Chapter 6 Trigonometric Identities RD Sharma Class 10 Solutions Trigonometric Identities Exercise 6.1 Prove the following trigonometric identities : Question 1. (1 – cos 2 A) cosec 2 A = 1 Solution: (1 – cos 2 A) cosec 2 A = 1 L.H.S. = (1 – cos 2 A) cosec 2 A = sin 2 A cosec 2 A (∵ 1 – cos 2 A = sin 2 A) = (sin A cosec A) 2 = (l) 2 = 1 = R.H.S. = [cosec A sin A] 2 = (1) 2= 1 = R.H.S. (∵ sin A cosec A = 1 Question 3. tan 2 θ cos 2 θ = 1- cos 2 θ Solution: Question 4. Solution: Question 5. (sec 2 θ – 1) (cosec 2 θ – 1) = 1 Solution: Question 6. Solution: Question 7. Solution: Question 8. Solution: Question 9. Solution: Question 10. Solution: Question 11. Solution: Question 12. Solution: Question 13. Solution: Question 14. Solution: Question 15. Solution: Question 16. tan 2 θ – sin 2 θ = tan 2 θ sin 2 θ Solution: Question 17. (sec θ + cos θ ) (sec θ – cos θ ) = tan 2 θ + sin 2 θ Solution: Question 18. (cosec θ + sin θ) (cosec θ – sin θ) = cot 2 θ + cos 2 θ Solution: Question 19. sec A (1 – sin A) (sec A + tan A) = 1 (C.B.S.E. 1993) Solution: Question 20. (cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1 Solution: Question 21. (1 + tan 2 θ) (1 – sin θ) (1 + sin θ) = 1 Solution: Question 22. sin 2 A cot 2 A + cos 2 A tan 2 A = 1 (C.B.S.E. 1992C) Solution: Question 23. Solution: Question 24. Solution: Question 25. Solution: Question 26. Solution: Question 27. Solution: Question 28. Solution: Question 29. Solution: Question 30. Solution: Question...

RD Sharma Class 10 Solutions Chapter 6 Trigonometric Identities RD Sharma Class 10 Solutions Trigonometric Identities Exercise 6.1 Prove the following trigonometric identities : Question 1. (1 – cos 2 A) cosec 2 A = 1 Solution: (1 – cos 2 A) cosec 2 A = 1 L.H.S. = (1 – cos 2 A) cosec 2 A = sin 2 A cosec 2 A (∵ 1 – cos 2 A = sin 2 A) = (sin A cosec A) 2 = (l) 2 = 1 = R.H.S. = [cosec A sin A] 2 = (1) 2= 1 = R.H.S. (∵ sin A cosec A = 1 Question 3. tan 2 θ cos 2 θ = 1- cos 2 θ Solution: Question 4. Solution: Question 5. (sec 2 θ – 1) (cosec 2 θ – 1) = 1 Solution: Question 6. Solution: Question 7. Solution: Question 8. Solution: Question 9. Solution: Question 10. Solution: Question 11. Solution: Question 12. Solution: Question 13. Solution: Question 14. Solution: Question 15. Solution: Question 16. tan 2 θ – sin 2 θ = tan 2 θ sin 2 θ Solution: Question 17. (sec θ + cos θ ) (sec θ – cos θ ) = tan 2 θ + sin 2 θ Solution: Question 18. (cosec θ + sin θ) (cosec θ – sin θ) = cot 2 θ + cos 2 θ Solution: Question 19. sec A (1 – sin A) (sec A + tan A) = 1 (C.B.S.E. 1993) Solution: Question 20. (cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1 Solution: Question 21. (1 + tan 2 θ) (1 – sin θ) (1 + sin θ) = 1 Solution: Question 22. sin 2 A cot 2 A + cos 2 A tan 2 A = 1 (C.B.S.E. 1992C) Solution: Question 23. Solution: Question 24. Solution: Question 25. Solution: Question 26. Solution: Question 27. Solution: Question 28. Solution: Question 29. Solution: Question 30. Solution: Question...

Answer: Answer: OR Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Question 14: Evaluate without using trigonometric tables: Answer: Answer: Question 16: Without using trigonometric tables, evaluate Answer: Question 17: Prove the identity: Answer: Question 18: If , show that Answer: Therefore LHS = RHS. Hence proved. Answer: Answer: Answer: Answer: Answer: Question 24: Without using trigonometric tables evaluate : Answer: Given: Question 25: Without using trigonometric tables evaluate : Answer: Given: