Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Exercise 21(B)

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(B)

Question 1. Prove that:

Solution:

(i)

– Hence Proved

(ii) Taking LHS,

– Hence Proved

(iii)

– Hence Proved

(iv)

– Hence Proved

(v) Taking LHS,

2 sin2 A + cos2 A

= 2 sin2 A + (1 – sin2 A)2

= 2 sin2 A+ 1 + sin4 A – 2 sin2 A

= 1 + sin4 A = RHS

– Hence Proved

(vi)

– Hence Proved

(vii)

– Hence Proved

(viii)

– Hence Proved

(ix)

– Hence Proved

Question 2. If x cos A + y sin A = m and x sin A – y cos A = n, then prove that:

x2 + y2 = m2 + n2 

Solution:

Taking RHS,

m2 + n2

= (x cos A + y sin A)2 + (x sin A – y cos A)2

= x2 cos2 A + y2 sin2 A + 2xy cos A sin A + x2 sin2 A + y2 cos2 A – 2xy sin A cos A

= x2 (cos2 A + sin2 A) + y2 (sin2 A + cos2 A)

= x2 + y2 [Since, cos2 A + sin2 A = 1]

= RHS

Question 3. If m = a sec A + b tan A and n = a tan A + b sec A, prove that m2 – n2 = a2 – b2

Solution:

Taking LHS,

m2 – n2

= (a sec A + b tan A)2 – (a tan A + b sec A)2

= a2 sec2 A + b2 tan2 A + 2 ab sec A tan A – a2 tan2 A – b2 sec2 A – 2ab tan A sec A

= a2 (sec2 A – tan2 A) + b2 (tan2 A – sec2 A)

= a2 (1) + b2 (-1) [Since, sec2 A – tan2 A = 1]

= a2 – b2

= RHS

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

Exercise 21(E) Solutions

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