RS Aggarwal Solutions Chapter 8 Trigonometric Identities Exercise 8B Class 10 Maths
Chapter Name | RS Aggarwal Chapter 8 Trigonometric Identities |
Book Name | RS Aggarwal Mathematics for Class 10 |
Other Exercises |
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Related Study | NCERT Solutions for Class 10 Maths |
Exercise 8B Solutions
1. If a cos θ + b sin θ m and a sin θ – b cos θ = n, prove that, (m2 + n2) = (a2 + b2)
Solution
We have m2 + n2 = [(a cos θ + b sin θ)2 + (a sin θ – b sin θ)2]
= (a2 cos2 θ + b2 sin2 θ + 2ab cos θ sin θ) + (a2 sin2 θ + b2 cos2 θ – 2abcos θ sin θ)
= a2 cos2 θ + b2 sin2 θ + a2 sin2 θ + b2 cos2 θ
= (a2 cos2 θ + b2 sin2 θ) + (b2 cos2 θ + b2 sin2 θ)
= a2(cos2 θ + sin2 θ) + b2 (cos2 θ + sin2 θ)
= a2 + b2 [∵ sin2 + cos2 = 1]
Hence, m2 + n2 = a2 + b2
2. If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x2 – y2) = (a2 – b2).
Solution
We have x2 – y2 = [(a sec θ + b tan θ)2 – (a tan θ + b sec θ)2]
= (a2 sec2 θ + b2 tan2 θ + 2ab sec θ tan θ) – (a2 tan2 θ + b2 sec2 θ + 2 ab tan θ sec θ)
= a2 sec2 θ + b2 tan2 θ – a2 tan2 θ – b2 sec2 θ
= (a2 sec2 θ – a2 tan2 θ) – (b2 sec2 θ – b2 tan2 θ)
= a2(sec2 θ – tan2 θ) – b2(sec2 θ- tan2 θ)
= a2 – b2 [∵ sec2 θ – tan2 θ = 1]
Hence, x2 – y2 = a2 – b2
3. If (x/a sin θ – y/b cos θ) = 1 and (x/a cos θ + y/b sin θ) = 1 prove that (x2/a2 + y2/b2) = 2.
Solution
We have:
(x/a sin θ – y/b cos θ) = 1
Squaring both side, we have:
(x/a sin θ – y/b cos θ)2 = (1)2
⇒ (x2/a2 sin2 θ + y2/b2 cos2 θ – 2.x/a × y/b sin θ cos θ) = 1 ...(i)
Again, (x/a cos θ + y/b sin θ) = 1
Squaring both side, we get:
(x/a cos θ + y/b sin θ)2 = 12
⇒ (x2/a2 cos2 θ + y2/b2 sin2 θ + 2.x/a × y/a sin θ cos θ) = 12 ....(ii)
Now adding (i) and (ii), we get:
(x2/a2 sin2 θ + y2/b2 cos2 θ – 2.x/a × y/b sin θ cos θ) + (x2/a2 cos2 θ + y2/b2 sin2 θ + 2.x/a × y/b sin θ cos θ)
⇒ x2/a2 sin2 θ + y2/b2 cos2 θ + x2/a2 cos2 θ + y2/b2 sin2 θ = 2
⇒ (x2/a2 sin2 θ + x2/a2 cos2 θ) + (y2/b2 cos2 θ + y2/b2 sin2 θ) = 2
⇒ x2/a2 (sin2 θ + cos2 θ) + y2/b2 (cos2 θ + sin2 θ) = 2
⇒ x2/a2 + y2/b2 = 2 [∵ sin2 θ + cos2 θ = 1]
∴ x2/a2 + y2/b2 = 2
4. If (sec θ + tan θ) = m and (sec θ – tan θ) = n, show that mn = 1
Solution
We have (sec θ + tan θ) = m ...(i)
Again, (sec θ – tan θ) = n ...(ii)
Now, multiplying (i) and (ii), we get:
(sec θ + tan θ) × (sec θ – tan θ) = mn
⇒ sec2 θ – tan2 θ = mn
⇒ 1 = mn [∵ sec2 θ – tan2 θ = 1]
∴ mn = 1
5. If (cosec θ + cot θ) = m and (cosec θ – cot θ) = n, show that mn = 1.
Solution
we have (cosec θ + cot θ) = m ...(i)
Again, (cosec θ – cot θ) = n ...(ii)
Now, multiplying (i) and (ii), we get:
(cosec θ + cot θ) × (cosec θ – cot θ) = mn
⇒ 1 = mn [∵ cosec2 θ – cot2 θ = 1]
∴ mn = 1
6. If x = a cos3 θ and y = b sin3 θ, Prove that (x/a)2/3 + (y/b)2/3 = 1.
Solution
We have x = a cos3 θ
⇒ x/a = cos3 θ ...(i)
Again, y = b sin3 θ
⇒ y/b = sin3 θ ...(ii)
Now, LHS = (x/a)2/3 + (y/b)2/3
= (cos3 θ)2/3 + (sin3 θ)2/3 [From (i) and (ii)]
= cos2 θ + sin2 θ
= 1
Hence, LHS = RHS
7. If (tan θ + sin θ) = m and (tan θ – sin θ) = n, prove that (m2 – n2)2 = 16 mn.
Solution
We have (tan θ + sin θ) = m and (tan θ – sin θ) = n
Now, LHS = (m2 – n2)2
= [tan θ + sin θ)2 – (tan θ – sin θ)2]2
= [(tan2 θ + sin2 θ + 2 tan θ sin θ) – (tan2 θ + sin2 θ – 2 tan θ sin θ)]2
= [(tan2 θ + sin2 θ + 2 tan θ sin θ – tan2 θ – sin2 θ + 2 tan θ sin θ)]2
= (4 tan θ sin θ)2
= 16 tan2 θ sin2 θ
= 16 (sin2 θ/cos2 θ) sin2 θ
= 16{(1 – cos2 θ)sin2 θ}/cos2 θ
= 16[tan2 θ (1 – cos2 θ)]
= 16(tan2 θ – tan2 θ cos2 θ)
= 16(tan2 θ – sin2 θ/cos2 θ × cos2 θ)s
= 16(tan2 θ – sin2 θ)
= 16 (tan θ + sin θ)(tan θ – sin θ)
= 16 mn [(tan θ + sin θ)(tan θ – sin θ) = mn]
∴ (m2 – n2)(m2 – n2)2 = 16 mn
8. If (cot θ + tan θ) = m and (sec θ – cos θ) = n prove that (m2n)2/3 – (mn2)2/3 = 1
Solution
We have (cot θ + tan θ) = m and (sec θ – cos θ) = n
Now, m2n = [(cot θ + tan θ)2 (sec θ – cos θ)]
∴ (m2n)2/3 = (sec3 θ)2/3 = sec2 θ
Again,
mn2 = [(cot θ + tan θ)(sec θ – cos θ)2]
9. If (cosec θ – sin θ) = a3 and (sec θ – cos θ) = b3, prove that a2b2 (a2 + b2) = 1
Solution
We have (cosec θ – sin θ) = a3
⇒ a3 = (1/sin θ – sin θ)
⇒ a3 = (1 – sin2 θ)/sin θ = cos2 θ/sin θ
∴ a = (cos2/3 θ)/(sin1/3 θ)
Again, (sec θ – cos θ) = b3
⇒ b3 = (1/cos θ – cos θ)
= (1 – cos2 θ)/cos θ
= (sin2 θ)/cos θ)
∴ b = (sin2/3 θ)/(cos1/3 θ)
Now, LHS = a2b2 (a2 + b2)
= a4b2 + a2b4
= a3(ab2) + (a2b2)b3
= cos2 θ + sin2 θ
= 1
= RHS
Hence, proved.
10. If (2 sin θ + 3 cos θ) = 2, prove that (3 sin θ – 2 cos θ) = ± 3
Solution
Given, (2 sin θ + 3 cos θ) = 2 ...(i)
We have (2 sin θ + 3 cos θ)2 + (3 sin θ – 2 cos θ)2
= 4 sin2 θ + 9 cos2 θ + 12 sin θ cos θ + 9 sin2 θ + 4 cos2 θ – 12 sin θ cos θ
= 4(sin2 θ + cos2 θ) + 9(sin2 θ + cos2 θ)
= 4 + 9
= 13
i.e., (2 sin θ + 3 cos θ)2 + (3 sin θ 2 cos θ)2 = 13
⇒ 22 + (3 sin θ – 2 cos θ)2 = 13
⇒ (3 sin θ – 2 cos θ)2 = 13 – 4
⇒ (3 sin θ – 2 cos θ)2 = 9
⇒ (3 sin θ – 2 cos θ) = ± 3
11. If (sin θ + cos θ) = √2, prove that cot θ = (√2 + 1).
Solution
We have, (sin θ + cos θ) = √2 cos θ
Dividing both sides by sin θ, we get
sin θ/sin θ + cos θ/sin θ = (√2 cos θ)/sin θ
⇒ 1 + cot θ = √2 cot θ
⇒ √2 cot θ – cot θ = 1
⇒ (√2 – 1) cot θ = 1
⇒ cot θ = 1/(√2 – 1)
⇒ cot θ = 1/(√2 – 1) × (√2 + 1)/(√2 + 1)
⇒ cot θ = (√2 + 1)/(2 – 1)
⇒ cot θ = (√2 + 1)/1
∴ cot θ = (√2 + 1)
12. If (cos θ + sin θ) = √2 sin θ, prove that (sin θ – cos θ) = √2 cos θ.
Solution
Given: cos θ + sin θ = √2 sin θ
We have (sin θ + cos θ)2 + (sin θ – cos θ)2 = 2(sin2 θ + cos2 θ)
⇒ (√2 sin θ)2 + (sin θ – cos θ)2 = 2
⇒ 2 sin2 θ + (sin θ – cos θ)2 = 2
⇒ (sin θ – cos θ)2 = 2 – 2 sin2 θ
⇒ (sin θ – cos θ)2 = 2(1 – sin2 θ)
⇒ (sin θ – cos θ)2 = 2 cos2 θ
⇒ (sin θ – cos θ) = √2 cos θ
Hence proved.
13. If sec θ + tan θ = p, prove that
(i) sec θ = 1/2(p + 1/p)
(ii) tan θ = 1/2(p – 1/p)
(iii) sin θ = (p2 – 1)/(p2 + 1)
Solution
(i) We have, sec θ + tan θ = p ...(1)
⇒ (sec θ + tan θ)/1 × (sec θ – tan θ)/(sec θ – tan θ) = p
⇒ (sec2 θ – tan2 θ)/(sec θ – tan θ) = p
⇒ 1/(sec θ – tan θ) = p
⇒ (sec θ – tan θ) = 1/p ...(2)
Adding (1) and (2), we get
2 sec θ = p + 1/p
⇒ sec θ = 1/2(p + 1/p)
(ii) Subtracting (2) from (1), we get
2 tan θ = (p – 1/p)
⇒ tan θ = ½(p – 1/p)
(iii) Using (i) and (ii), we get
sin θ = tan θ /sec θ
∴ sin θ = (p2 – 1)/(p2 + 1)
14. If tan A = n tan B and sin A = m sin B, prove that cos2 A = (m2 – 1)/(n2 – 1).
Solution
We have tan A = n tan B
⇒ cot B = n/tan A ...(i)
Again, sin A = m sin B
⇒ cosec B = m/sin A ...(ii)
Squaring (i) and (ii) and subtracting (ii) from (i), we get
⇒ m2/sin2 A – n2/tan2 A = cosec2 B – cot2 B
⇒ m2/sin2 A – n2 cos/sin2 A = 1
⇒ m2 – n2 cos2 A = sin2 A
⇒ m2 – n2 cos2 A = 1 - cos2 A
⇒ n2 cos2 A – cos2 A = m2 – 1
⇒ cos2 A(n2 – 1) = (m2 – 1)
⇒ cos2 A = (m2 – 1)/(n2 – 1)
∴ cos2 A = (m2 – 1)/(n2 – 1)
15. If m = (cos θ - sin θ) and n = (cos θ + sin θ) then show that √m/n + √n/m = 2/√(1-tan2θ)
Solution
