Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1

Students can download Maths Chapter 6 Trigonometry Ex 6.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1

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
= \(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}\)
= \(\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}\)
[cos2 θ + sin2 θ = 1]
= \(\frac{1}{\sin \theta \cos \theta}\)
= sec θ . cosec θ = R. H. S
∴ cot θ + tan θ = sec θ cosec θ

Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1

(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) \(\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}\) = tan2 θ
(ii) \(\frac{\cos \theta}{1+\sin \theta}\) = sec θ – tan θ
Answer:
(i) \(\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}\) = tan2 θ
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 1
(ii) \(\frac{\cos \theta}{1+\sin \theta}\) = sec θ – tan θ
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 2
Aliter:
L.H.S = \(\frac{\cos \theta}{1-\sin \theta}\)
[conjugate (1 – sin θ)]
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 3

Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 3.
Prove the following identities.
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 4
Solution:
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 5
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 6
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 7
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 77

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)
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 8
= 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

Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 5.
Prove the following identities.
(i) sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
(ii) \(\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}=\frac{\csc \theta-1}{\csc \theta+1}\)
Answer:
(i) L.H.S = sec4 θ (1 – sin4 θ) – 2 tan2 θ
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 9
L.H.S = R.H.S
∴ sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1

(ii) \(\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}=\frac{\csc \theta-1}{\csc \theta+1}\)
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 10
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 11
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 111

Question 6.
Prove the following identities.
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 12
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 13
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 14
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 15

Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 7.
(i) If sin θ + cos θ = \(\sqrt { 3 }\), then prove that tan θ + cot θ = 1.
(ii) If \(\sqrt { 3 }\) sin θ – cos θ = θ, then show that tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)
Answer:
sin θ + cos θ = \(\sqrt { 3 }\) (squaring on both sides)
(sin θ + cos θ)2 = (\(\sqrt { 3 }\))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
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 16
L.H.S = R.H.S ⇒ tan θ + cot θ = 1

(ii) If \(\sqrt { 3 }\) sin θ – cos θ = 0
To prove tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)
\(\sqrt { 3 }\) sin θ – cos θ = 0
\(\sqrt { 3 }\) sin θ = cos θ
\(\frac{\sin \theta}{\cos \theta}=\frac{1}{\sqrt{3}}\)
tan θ = tan 30°
θ = 30°
L.H.S = tan 3θ°
= tan3 (30°)
= tan 90°
= undefined (∝)
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 22
∴ tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)

Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 8.
(i) If \(\frac{\cos \alpha}{\cos \beta}=m\) and \(\frac{\cos \alpha}{\cos \beta}=n\) 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
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 17
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 = \(\frac{1}{\tan \theta}\) + tan θ
= \(\frac{1+\tan ^{2} \theta}{\tan \theta}\) = \(\frac{\sec ^{2} \theta}{\tan \theta}\)
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 18
y = sec θ – cos θ
= \(\frac{1}{\cos \theta}-\cos \theta=\frac{1-\cos ^{2} \theta}{\cos \theta}\)
y = \(\frac{\sin ^{2} \theta}{\cos \theta}\)
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 19

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
= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\)
L.H.S = q(p2 – 1)
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 20

(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

Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 10.
If \(\frac{\cos \theta}{1+\sin \theta}\) = \(\frac { 1 }{ a } \), then prove that \(\frac{a^{2}-1}{a^{2}+1}\) = sin θ
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 21
Samacheer Kalvi 10th Maths Guide Chapter 6 Trigonometry Ex 6.1 223

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