You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5

Question 1.
Find the value of cos 2A, A lies in the first quadrant, when
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 1
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 2

Question 2.
If θ is an acute angle, then find
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 3
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 4
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 5

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 6
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 7
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 8

Question 4.
Prove that cos 5 θ = 16 cos5 θ – 20 cos3 θ + 5 cos θ.
Solution:
cos 5 θ = cos(2θ + 3θ) = cos 2θ cos 3θ – sin 2θ sin 3θ
= (2 cos2 θ – 1) (4 cos3 θ – 3 cos θ) – 2 sin θ cos θ (3 sin θ – 4 sin3 θ)
= 8cos5 θ – 6 cos3 θ – 4 cos3 θ + 3 cos θ – 6 sin2 θ cos θ + 8 cos θ sin4 θ
= 8 cos5 θ – 6 cos3 θ – 4 cos3 θ + 3 cos θ – 6(1 – cos2 θ) cos θ + 8 cos θ (1 – cos2 θ)2
= 8 cos5 θ – 6 cos3 θ – 4 cos3 θ + 3 cos θ – 6 cos θ + 6 cos3 θ + 8 cos 0(1+ cos4 θ – 2 cos2 θ)
= 8 cos5 θ – 6 cos3 θ – 4 cos3 θ + 3 cos θ – 6 cos θ + 6 cos3 θ + 8 cos θ + 8 cos5 θ – 16 cos3 θ
= 16 cos5 θ – 20 cos3 θ + 5 cos θ = RHS

Question 5.
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Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 21

Question 6.
If A + B = 45°, show that (1 + tanA) (1 + tanB) = 2.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 22
Now LHS = (1 + tan A) (1 + tan B)
= tan A + tan B + tan A tan B + 1
= (1 – tan A tan B) + (tan A tan B + 1) from (1)
= 2 = RHS

Question 7.
Prove that (1 + tan 1°)(1 + tan 2°)(1 + tan 3°)… (1 + tan 44°) is a multiple of 4.
Solution:
1 + tan 44° = 1 + tan (45° – 1°)
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(1 + tan 1°)(1 + tan 44°) = 2
Similarly (1 + tan 2°) (1 + tan 43°) = 2
(1 + tan 3°) (1 + tan 42°) = 2
(1 + tan 22°) (1 + tan 23°) = 2
= (1 + tan 1°) (1 + tan 2°)… (1 + tan 44°) = 2 × 2 × … 22 times
It is a multiple of 4.

Question 8.
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Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 26

Question 9.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5

Question 10.
Prove that (1 + sec 2θ)(1 + sec 4θ)….. (1 + sec 2nθ) = tan 2nθ
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 30

Question 11.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 Additional Questions

Question 1.
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Solution:
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Question 2.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5

Question 3.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 40

Question 4.
Show that 4 sin A sin (60° + A). sin(60° – A) = sin 3A
Solution:
LHS = 4sinAsin(60° + A). sin(60° – A)
= 4 sin A{sin (60° + A). sin (60° – A)}
= 4 sin A {sin2 60° – sin2 A)}
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 41

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