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

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(E)

Question 1. Prove the following identities:

Solution:

(i) Taking LHS,

1/ (cos A + sin A) + 1/ (cos A – sin A)

= RHS

– Hence Proved

(ii) Taking LHS, cosec A – cot A

= RHS

– Hence Proved

(iii) Taking LHS, 1 – sin2 A/ (1 + cos A)

= RHS

– Hence Proved

(iv) Taking LHS,

(1 – cos A)/ sin A + sin A/ (1 – cos A)

= RHS

– Hence Proved

(v) Taking LHS, cot A/ (1 – tan A) + tan A/ (1 – cot A)

= RHS

– Hence Proved

(vi) Taking LHS, cos A/ (1 + sin A) + tan A

= RHS

– Hence Proved

(vii) Consider LHS,

= (sin A/(1 – cos A)) – cot A

We know that, cot A = cos A/sin A

So,
= (sin2 A – cos A + cos2 A)/(1 – cos A) sin A

= (1 – cos A)/(1 – cos A) sin A

= 1/sin A

= cosec A

(viii) Taking LHS, (sin A – cos A + 1)/ (sin A + cos A – 1)

= RHS

– Hence Proved

(ix) Taking LHS,

= RHS

– Hence Proved

(x) Taking LHS,

= RHS

– Hence Proved

(xi) Taking LHS,

= RHS

– Hence Proved

(xii) Taking LHS,

= RHS

– Hence Proved

(xiii) Taking LHS,

= RHS

– Hence Proved

(xiv) Taking LHS,

= RHS

– Hence Proved

(xv) Taking LHS,

sec4 A (1 – sin4 A) – 2 tan2 A

= sec4 A(1 – sin2 A) (1 + sin2 A) – 2 tan2 A

= sec4 A(cos2 A) (1 + sin2 A) – 2 tan2 A

= sec2 A + sin2 A/ cos2 A – 2 tan2 A

= sec2 A – tan2 A

= 1 = RHS

– Hence Proved

(xvi) cosec4 A(1 – cos4 A) – 2 cot2 A

= cosec4 A (1 – cos2 A) (1 + cos2 A) – 2 cot2 A

= cosec4 A (sin2 A) (1 + cos2 A) – 2 cot2 A

= cosec2 A (1 + cos2 A) – 2 cot2 A

= cosec2 A + cos2 A/sin2 A – 2 cot2 A

= cosec2 A + cot2 A – 2 cot2 A

= cosec2 A – cot2 A

= 1 = RHS

– Hence Proved

(xvii) (1 + tan A + sec A) (1 + cot A – cosec A)

= 1 + cot A – cosec A + tan A + 1 – sec A + sec A + cosec A – cosec A sec A

= 2 + cos A/sin A+ sin A/cos A – 1/(sin A cos A)

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

= 2 + 1/(sin A cos A) – 1/(sin A cos A)

= 2 = RHS

– Hence Proved

Question 2. If sin A + cos A = p

and sec A + cosec A = q, then prove that: q(p2 – 1) = 2p

Solution:

Taking the LHS, we have

q(p2 – 1) = (sec A + cosec A) [(sin A + cos A)2 – 1]

= (sec A + cosec A) [sin2 A + cos2 A + 2 sin A cos A – 1]

= (sec A + cosec A) [1 + 2 sin A cos A – 1]

= (sec A + cosec A) [2 sin A cos A]

= 2sin A + 2 cos A

= 2p

Question 3. If x = a cos θ and y = b cot θ, show that:

a2/ x2 – b2/ y2 = 1

Solution:

Taking LHS,

a2/ x2 – b2/ y2

Question 4. If sec A + tan A = p, show that:

sin A = (p2 – 1)/ (p2 + 1)

Solution:

Taking RHS, (p2 – 1)/ (p2 + 1)

Question 5. If tan A = n tan B and sin A = m sin B, prove that:

cos2 A = m2 – 1/ n2 – 1

Solution:

Given,

tan A = n tan B

n = tan A/ tan B

And, sin A = m sin B

m = sin A/ sin B

Now, taking RHS and substitute for m and n

m2 – 1/ n2 – 1

Question 6. (i) If 2 sin A – 1 = 0, show that:

sin 3A = 3 sin A – 4 sin3 A

(ii) If 4 cos2 A – 3 = 0, show that:

cos 3A = 4 cos2 A – 3 cos A

Solution:

(i) Given, 2 sin A – 1 = 0

So, sin A = ½

We know, sin 30o = 1/2

Hence, A = 30o

Now, taking LHS

sin 3A = sin 3(30o) = sin 30o = 1

RHS = 3 sin 30o – 4 sin3 30o = 3 (1/2) – 4 (1/2)3 = 3 – 4(1/8) = 3/2 – ½ = 1

Therefore, LHS = RHS

(ii) Given, 4 cos2 A – 3 = 0

4 cos2 A = 3

cos2 A = 3/4

cos A = √3/2

We know, cos 30o = √3/2

Hence, A = 30o

Now, taking

LHS = cos 3A = cos 3(30o) = cos 90o = 0

RHS = 4 cos3 A – 3 cos A = 4 cos3 30o – 3 cos 30o = 4 (√3/2)3 – 3 (√3/2)

= 4 (3√3/8) – 3√3/2

= 3√3/2 – 3√3/2

= 0

Therefore, LHS = RHS

Question 7. Evaluate:

Solution:

(i)

= 2 (1)2 + 12 – 3

= 2 + 1 – 3 = 0

(ii)

= 1 + 1 = 2

(iii)

(iv) cos 40o cosec 50o + sin 50o sec 40o

= cos (90 – 50)o cosec 50o + sin (90 – 50)o sec 40o

= sin 50o cosec 50o + cos 40o sec 40o

= 1 + 1 = 2

(v) sin 27o sin 63o – cos 63o cos 27o

= sin (90 – 63)o sin 63o – cos 63o cos (90 – 63)o

= cos 63o sin 63o – cos 63o sin 63o

= 0

(vi)

(vii) 3 cos 80o cosec 10o + 2 cos 59o cosec 31o

= 3 cos (90 – 10)o cosec 10o + 2 cos (90 – 31)o cosec 31o

= 3 sin 10o cosec 10o + 2 sin 31o cosec 31o

= 3 + 2 = 5

(viii)

Question 8. Prove that:

(i) tan (55o + x) = cot (35o – x)

(ii) sec (70o – θ) = cosec (20o + θ)

(iii) sin (28o + A) = cos (62o – A)

(iv) 1/ (1 + cos (90o – A)) + 1/(1 – cos (90o – A)) = 2 cosec2 (90o – A)

(v) 1/ (1 + sin (90o – A)) + 1/(1 – sin (90o – A)) = 2 sec2 (90o – A)

Solution:

(i) tan (55o + x) = tan [90o – (35o – x)] = cot (35o – x)

(ii) sec (70o – θ) = sec [90o – (20o + θ)] = cosec (20o + θ)

(iii) sin (28o + A) = sin [90o – (62o – A)] = cos (62o – A)

(iv)

(v)

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

Exercise 21(E) Solutions

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