Selina Solutions Class 10 Maths Chapter 21 Trigonometrical Identities

 

Selina Solutions Class 10 Maths Chapter 21 Trigonometrical Identities

The science related to the measurement of triangles is called trigonometry. Trigonometric ratios and its relations are used to prove trigonometric identities. In addition, trigonometric ratios of complementary angles and the use of trigonometric tables are other topics covered in this chapter. Since this chapter lays the foundation for high grade mathematics, students should gain a strong grip on this chapter. For this purpose, www.mathspdfsolution.co  has created Selina Solutions for Class 10 Mathematics prepared by expert faculty with vast academic experience. It also improves students' problem-solving skills, which are important from the exam point of view. Selina Solutions for Class 10 Mathematics Chapter 21 Trigonometric Identification PDFs are available practice-wise in the link below.

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

Exercise 21(E) Solutions

Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities

Exercise 21(A) Page No: 237

Prove the following identities:

Question 1. sec A – 1/ sec A + 1 = 1 – cos A/ 1 + cos A

Solution:

– Hence Proved

Question 2. 1 + sin A/ 1 – sin A = cosec A + 1/ cosec A – 1

Solution:

– Hence Proved

Question 3. 1/ tan A + cot A = cos A sin A

Solution:

Taking L.H.S,

– Hence Proved

Question 4. tan A – cot A = 1 – 2 cos2 A/ sin A cos A

Solution:

Taking LHS,

– Hence Proved

Question 5. sin4 A – cos4 A = 2 sin2 A – 1

Solution:

Taking L.H.S,

sin4 A – cos4 A

= (sin2 A)2 – (cos2 A)2

= (sin2 A + cos2 A) (sin2 A – cos2 A)

= sin2A – cos2A

= sin2A – (1 – sin2A) [Since, cos2 A = 1 – sin2 A]

= 2sin2 A – 1

– Hence Proved

Question 6. (1 – tan A)2 + (1 + tan A)2 = 2 sec2 A

Solution:

Taking L.H.S,

(1 – tan A)2 + (1 + tan A)2

= (1 + tan2 A + 2 tan A) + (1 + tan2 A – 2 tan A)

= 2 (1 + tan2 A)

= 2 sec2 A [Since, 1 + tan2 A = sec2 A]

– Hence Proved

Question 7. cosec4 A – cosec2 A = cot4 A + cot2 A

Solution:

cosec4 A – cosec2 A

= cosec2 A(cosec2 A – 1)

= (1 + cot2 A) (1 + cot2 A – 1)

= (1 + cot2 A) cot2 A

= cot4 A + cot2 A = R.H.S

– Hence Proved

Question 8. sec A (1 – sin A) (sec A + tan A) = 1

Solution:

Taking L.H.S,

sec A (1 – sin A) (sec A + tan A)

– Hence Proved

Question 9. cosec A (1 + cos A) (cosec A – cot A) = 1

Solution:

Taking L.H.S,

– Hence Proved

Question 10. sec2 A + cosec2 A = sec2 A . cosec2 A

Solution:

Taking L.H.S,

– Hence Proved

Question 11. (1 + tan2 A) cot A/ cosec2 A = tan A

Solution:

Taking L.H.S,

= RHS

– Hence Proved

Question 12. tan2 A – sin2 A = tan2 A. sin2 A

Solution:

Taking L.H.S,

tan2 A – sin2 A

– Hence Proved

Question 13. cot2 A – cos2 A = cos2A. cot2A

Solution:

Taking L.H.S,

cot2 A – cos2 A

– Hence Proved

Question 14. (cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A

Solution:

Taking L.H.S,

(cosec A + sin A) (cosec A – sin A)

= cosec2 A – sin2 A

= (1 + cot2 A) – (1 – cos2 A)

= cot2 A + cos2 A = R.H.S

– Hence Proved

Question 15. (sec A – cos A)(sec A + cos A) = sin2 A + tan2 A

Solution:

Taking L.H.S,

(sec A – cos A)(sec A + cos A)

= (sec2 A – cos2 A)

= (1 + tan2 A) – (1 – sin2 A)

= sin2 A + tan2 A = RHS

– Hence Proved

Question 16. (cos A + sin A)2 + (cosA – sin A)2 = 2

Solution:

Taking L.H.S,

(cos A + sin A)2 + (cosA – sin A)2

= cos2 A + sin2 A + 2cos A sin A + cos2 A – 2cosA.sinA

= 2 (cos2 A + sin2 A) = 2 = R.H.S

– Hence Proved

Question 17. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1

Solution:

Taking LHS,

(cosec A – sin A)(sec A – cos A)(tan A + cot A)

= RHS

– Hence Proved

Question 18. 1/ sec A + tan A = sec A – tan A

Solution:

Taking LHS,

= RHS

– Hence Proved

Question 19. cosec A + cot A = 1/ cosec A – cot A

Solution:

Taking LHS,

cosec A + cot A

= RHS

– Hence Proved

20. sec A – tan A/ sec A + tan A = 1 – 2 secA tanA + 2 tan2 A

Solution:

Taking LHS,

= 1 + tan2 A + tan2 A – 2 sec A tan A

= 1 – 2 sec A tan A + 2 tan2 A = RHS

– Hence Proved

Question 21. (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

Solution:

Taking LHS,

(sin A + cosec A)2 + (cos A + sec A)2

= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2cos A sec A

= (sin2 A + cos2 A ) + cosec2 A + sec2 A + 2 + 2

= 1 + cosec2 A + sec2 A + 4

= 5 + (1 + cot2 A) + (1 + tan2 A)

= 7 + tan2 A + cot2 A = RHS

– Hence Proved

Question 22. sec2 A. cosec2 A = tan2 A + cot2 A + 2

Solution:

Taking,

RHS = tan2 A + cot2 A + 2 = tan2 A + cot2 A + 2 tan A. cot A

= (tan A + cot A)2 = (sin A/cos A + cos A/ sin A)2

= (sin2 A + cos2 A/ sin A.cos A)2 = 1/ cos2 A. sin2 A

= sec2 A. cosec2 A = LHS

– Hence Proved

Question 23. 1/ 1 + cos A + 1/ 1 – cos A = 2 cosec2 A

Solution:

Taking LHS,

= RHS

– Hence Proved

Question 24. 1/ 1 – sin A + 1/ 1 + sin A = 2 sec2 A

Solution:

Taking LHS,

= RHS

– Hence Proved

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

Exercise 21(E) Solutions

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