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

This exercise contains problems based on all the concepts discussed in the chapter. Students solving these problems will acquire strong conceptual knowledge. The Selina Solutions for Class 10 Maths is the right place for students to clear their doubts and also learn the right steps to solve problems. The solutions to the Concise Selina Solutions for Class 10 Maths Chapter 21 Trigonometrical Identities Exercise 21(E) are available in a PDF, provided in the links enclosed below.

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

 

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Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

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

1. Prove the following identities:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 1

Solution:

(i) Taking LHS,

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 2

= RHS

– Hence Proved

(ii) Taking LHS, cosec A – cot A

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 3

= RHS

– Hence Proved

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 4

= RHS

– Hence Proved

(iv) Taking LHS,

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 5

= RHS

– Hence Proved

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 6

= RHS

– Hence Proved

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 7

= 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)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 8

= RHS

– Hence Proved

(ix) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 9

= RHS

– Hence Proved

(x) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 10

= RHS

– Hence Proved

(xi) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 11

= RHS

– Hence Proved

(xii) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 12

= RHS

– Hence Proved

(xiii) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 13

= RHS

– Hence Proved

(xiv) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 14

= 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

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

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 15

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

sin A = (p2 – 1)/ (p2 + 1)

Solution:

Taking RHS, (p2 – 1)/ (p2 + 1)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 25

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 16

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

7. Evaluate:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 17

Solution:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 18(i)

= 2 (1)2 + 12 – 3

= 2 + 1 – 3 = 0

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 19(ii)

= 1 + 1 = 2

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 20(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)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 21

(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)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 22

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

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

(iv)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 23

(v)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 24

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