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

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

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

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

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

sin A = (p² – 1)/ (p² + 1)

Solution:

Taking RHS, (p² – 1)/ (p² + 1)

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

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 30^o = 1/2

Hence, A = 30^o

Now, taking LHS

sin 3A = sin 3(30^o) = sin 30^o = 1

RHS = 3 sin 30^o – 4 sin³ 30^o = 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 30^o = √3/2

Hence, A = 30^o

Now, taking

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

RHS = 4 cos³ A – 3 cos A = 4 cos³ 30^o – 3 cos 30^o = 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:

Solution:

(i)

= 2 (1)² + 1² – 3

= 2 + 1 – 3 = 0

(ii)

= 1 + 1 = 2

(iii)

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

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

= sin 50^o cosec 50^o + cos 40^o sec 40^o

= 1 + 1 = 2

(v) sin 27^o sin 63^o – cos 63^o cos 27^o

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

= cos 63^o sin 63^o – cos 63^o sin 63^o

= 0

(vi)

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

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

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

= 3 + 2 = 5

(viii)

8. Prove that:

(i) tan (55^o + x) = cot (35^o – x)

(ii) sec (70^o – θ) = cosec (20^o + θ)

(iii) sin (28^o + A) = cos (62^o – A)

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

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

Solution:

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

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

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

(iv)

(v)