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## 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

Solution:

Question 2.

If θ is an acute angle, then find

Solution:

Question 3.

Solution:

Question 4.

Prove that cos ^{5 }θ = 16 cos^{5} θ – 20 cos^{3} θ + 5 cos θ.

Solution:

cos ^{5 }θ = cos(2θ + 3θ) = cos 2θ cos 3θ – sin 2θ sin 3θ

= (2 cos^{2} θ – 1) (4 cos^{3} θ – 3 cos θ) – 2 sin θ cos θ (3 sin θ – 4 sin^{3} θ)

= 8cos^{5} θ – 6 cos^{3} θ – 4 cos^{3} θ + 3 cos θ – 6 sin^{2} θ cos θ + 8 cos θ sin^{4} θ

= 8 cos^{5} θ – 6 cos^{3} θ – 4 cos^{3} θ + 3 cos θ – 6(1 – cos^{2} θ) cos θ + 8 cos θ (1 – cos^{2} θ)^{2}

= 8 cos^{5} θ – 6 cos^{3} θ – 4 cos^{3} θ + 3 cos θ – 6 cos θ + 6 cos^{3} θ + 8 cos 0(1+ cos^{4} θ – 2 cos^{2} θ)

= 8 cos^{5} θ – 6 cos^{3} θ – 4 cos^{3} θ + 3 cos θ – 6 cos θ + 6 cos^{3} θ + 8 cos θ + 8 cos^{5} θ – 16 cos^{3} θ

= 16 cos^{5} θ – 20 cos^{3} θ + 5 cos θ = RHS

Question 5.

Solution:

Question 6.

If A + B = 45°, show that (1 + tanA) (1 + tanB) = 2.

Solution:

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°)

(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.

Solution:

Question 9.

Solution:

Question 10.

Prove that (1 + sec 2θ)(1 + sec 4θ)….. (1 + sec 2^{n}θ) = tan 2^{n}θ

Solution:

Question 11.

Solution:

### Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

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 {sin^{2} 60° – sin^{2} A)}