Trigonometry Ex 6.1-10th Std Maths-Book Back Question And Answer
Question 1.
Prove the following identities.
(i) cot θ + tan θ = sec θ cosec θ
(ii) tan4 θ + tan2 θ = sec4 θ – sec2 θ
Answer:
(i) L. H. S = cot θ + tan θ
= cosθsinθ+sinθcosθ
= cos2θ+sin2θsinθcosθ
[cos2 θ + sin2 θ = 1]
= 1sinθcosθ
= sec θ . cosec θ = R. H. S
∴ cot θ + tan θ = sec θ cosec θ
(ii) tan4 θ + tan2 θ = sec4 θ – sec2 θ
L.H.S = tan4 θ + tan2 θ
= tan2 θ (tan2 θ + 1)
= tan2 θ sec2 θ
R.H.S = sec4 θ – sec2 θ
= sec2 θ (sec2 θ – 1)
= sec2 θ tan2 θ
L.H.S = R.H.S
∴ tan4 θ + tan2 θ = sec4 θ – sec2 θ
Question 2.
Prove the following identities.
(i) 1−tan2θcot2θ−1 = tan2 θ
(ii) cosθ1+sinθ = sec θ – tan θ
Answer:
(i) 1−tan2θcot2θ−1 = tan2 θ
(ii) cosθ1+sinθ = sec θ – tan θ
Aliter:
L.H.S = cosθ1−sinθ
[conjugate (1 – sin θ)]
Question 3.
Prove the following identities.
Solution:
Question 4.
Prove the following identities.
(i) sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
(ii) (sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
Answer:
(i) sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
L.H.S = sec6 θ
= (sec2 θ)3 = (1 + tan2 θ)3
= 1 + (tan2 θ)3 + 3 (1) (tan2 θ) (1 + tan2 θ) [(a + b)3 = a3 + b3 + 3 ab (a + b)]
= 1 + tan6 θ + 3 tan2 θ(1 + tan2 θ)
= 1 + tan6 θ + 3 tan2 θ (sec2 θ)
= 1 + tan6 θ + 3 tan2 θ sec2 θ
= tan6 θ + 3 tan2 θ sec2 θ + 1
L.H.S = R.H.S
(ii) (sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
L.H.S = (sin θ + sec θ)2 + (cos θ + cosec θ)2]
= [sin2 θ + sec2 θ + 2 sin θ sec θ + cos2 θ + cosec2 θ + 2 cos θ cosec θ]
= (sin2 θ + cos2 θ) + (sec2 θ + cosec2 θ) + 2 (sin θ sec θ + cos θ cosec θ)
= 1 + sec2 θ + cosec2 θ + 2 sec θ cosec θ
= 1 + (sec θ + cosec θ)2
L.H.S = R.H.S
∴ (sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
Question 5.
Prove the following identities.
(i) sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
(ii) cotθ−cosθcotθ+cosθ=cscθ−1cscθ+1
Answer:
(i) L.H.S = sec4 θ (1 – sin4 θ) – 2 tan2 θ
L.H.S = R.H.S
∴ sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
(ii) cotθ−cosθcotθ+cosθ=cscθ−1cscθ+1
Question 6.
Prove the following identities.
Answer:
Question 7.
(i) If sin θ + cos θ = 3–√, then prove that tan θ + cot θ = 1.
(ii) If 3–√ sin θ – cos θ = θ, then show that tan 3θ = 3tanθ−tan3θ1−3tan2θ
Answer:
sin θ + cos θ = 3–√ (squaring on both sides)
(sin θ + cos θ)2 = (3–√)2
sin2 θ + cos2 θ + 2 sin θ cos θ = 3
1 + 2 sin θ cos θ = 3
2 sin θ cos θ = 3 – 1
2 sin θ cos θ = 2
∴ sin θ cos θ = 1
L.H.S = tan θ + cot θ
L.H.S = R.H.S ⇒ tan θ + cot θ = 1
(ii) If 3–√ sin θ – cos θ = 0
To prove tan 3θ = 3tanθ−tan3θ1−3tan2θ
3–√ sin θ – cos θ = 0
3–√ sin θ = cos θ
sinθcosθ=13√
tan θ = tan 30°
θ = 30°
L.H.S = tan 3θ°
= tan3 (30°)
= tan 90°
= undefined (∝)
∴ tan 3θ = 3tanθ−tan3θ1−3tan2θ
Question 8.
(i) If cosαcosβ=m and cosαcosβ=n then prove that (m2 + n2) cos2 β = n2
(ii) If cot θ + tan θ = x and sec θ – sec θ – cos θ = y, then prove that (x2y)2/3 – (xy2)2/3 = 1
Answer:
(i) L.H.S = (m2 + n2) cos2 β
L.H.S = R.H.S ⇒ ∴ (m2 + n2) cos2 β = n2
(ii) Given cot θ + tan θ = x sec θ – cos θ = y
x = cot θ + tan θ
x = 1tanθ + tan θ
= 1+tan2θtanθ = sec2θtanθ
y = sec θ – cos θ
= 1cosθ−cosθ=1−cos2θcosθ
y = sin2θcosθ
Question 9.
(i) If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q (p2 – 1) = 2 p
(ii) If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
Answer:
(i) p = sin θ + cos θ
p2 = (sin θ + cos θ)2
= sin2 θ + cos2 θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ
q = sec θ + cosec θ
= 1cosθ+1sinθ=sinθ+cosθsinθcosθ
L.H.S = q(p2 – 1)
(ii) sin θ (1 + sin2 θ) = cos2 θ
sin θ (1 + 1 – cos2 θ) = cos2 θ
sin θ (2 – cos2 θ) = cos2 θ
Squaring on both sides,
sin2 θ (2 – cos2 θ)2 = cos4 θ
(1 – cos2 θ) (4 + cos4 θ – 4 cos2 θ) = cos4 θ
4 cos4 θ – 4 cos2 θ – cos6 θ + 4 cos4 θ = cos4 θ
4 + 5 cos4 θ – 8 cos2 θ – cos6 θ = cos4 θ
– cos6 θ + 5 cos4 θ – cos4 θ – 8 cos2 θ = -4
– cos6 θ + 4 cos4 θ – 8 cos2 θ = -4
cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
Hence it is proved
Question 10.
If cosθ1+sinθ = 1a, then prove that a2−1a2+1 = sin θ
Answer:
