Exercise 4.2 - Chapter 4 - Trigonometry - 11th Business Maths Guide Samacheer Kalvi Solutions

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Exercise 4.2

Text Book Back Questions and Answers

Question 1.
Find the values of the following:
(i) cosec 15°
(ii) sin (-105°)
(iii) cot 75°
Solution:

Consider sin 15° = sin(45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°

(ii) sin (-105°) = -sin (105°) (∵ sin (-θ) = – sin θ)
= -[sin(60° + 45°)]
= -[sin 60° cos 45° + cos 60° sin 45°]

Consider tan 75° = tan (30° + 45°)

Question 2.
Find the values of the following:
(i) sin 76° cos 16° – cos 76° sin 16°

(iii) cos 70° cos 10° – sin 70° sin 10°
(iv) cos² 15° – sin² 15°
Solution:
(i) Given that, sin 76° cos 16° – cos 76° sin 16° (∴ This is of the form sin(A – B))
= sin(76° – 16°)
= sin 60°

(iii) Given that cos 70° cos 10° – sin 70° sin 10°
(This is of the form of cos (A + B), A = 70°, B = 10°)
= cos (70° + 10°)
= cos 80°

(iv) cos² 15° – sin² 15°
[∵ cos 2A = cos² A – sin² A, Here A = 15°]
= cos (2 × 15°)
= cos 30°

Question 3.

 find the values of the following:
(i) cos(A + B)
(ii) sin(A – B)
(iii) tan(A – B)
Solution:

[B lies in 3rd quadrant. tan B is positive.]
(i) cos(A + B) = cos A cos B – sin A sin B

(ii) sin(A – B) = sin A cos B – cos A sin B

(iii) tan(A – B)

Question 4.

Question 5.
Prove that 2 tan 80° = tan 85° – tan 5°.
Solution:
Consider tan 80° = tan(85° – 5°)

∴ 2 tan 80° = tan 85° – tan 5°
Hence Proved.

Question 6.

Question 7.
If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of

Solution:
Given A + B = 45°
tan (A + B) = tan 45°

tan A + tan B = 1 – tan A . tan B
tan A + tan B + tan A tan B = 1
Add 1 on both sides we get,
(1 + tan A) + tan B + tan A tan B = 2
1(1+ tan A) + tan B (1 + tan A) = 2
(1 + tan A) (1 + tan B) = 2 ……. (1)

Question 8.
Prove that
(i) sin(A + 60°) + sin(A – 60°) = sin A.
(ii) tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0
Solution:
(i) LHS = sin (A + 60°) + sin (A – 60°)
= sin A cos 60° + cos A sin 60° + sin A cos 60° – cos A sin 60°
= 2 sin A cos 60°

= sin A
= RHS

(ii) 4A = 3A + A
tan 4A = tan (3A + A)

on cross multiplication we get,
tan 3A + tan A = tan 4A (1 – tan 3A tan A) = tan 4A – tan 4A tan 3A tanA
i.e., tan 4A tan 3A tan A + tan 3A + tan A = tan 4A
(or) tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0

Question 9.
(i) If tan θ = 3 find tan 3θ
(ii) If sin A = 12/13, find sin 3A.
Solution:
(i) tan θ = 3

We know that sin 3A = 3 sin A – 4 sin³ A

Question 10.
If sin A = 3/5, find the values of cos 3A and tan 3A.
Solution:

Question 11.
Prove that

Solution:

Question 12.
If tan A – tan B = x and cot B – cot A = y prove that cot(A – B) 

Solution:

Hence proved.

Question 13.
If sin α + sin β = a and cos α + cos β = b, then prove that cos(α – β)

Solution:
Consider a² + b² = sin²α + sin²β + 2 sin α sin β + cos²α + cos²β + 2 cos α cos β
a² + b² = (sin²α + cos²α) + (sin²β + cos²β) + 2[cos α cos β + sin α sin β]
a² + b² = 1 + 1 + 2 cos(α – β)

Question 14.

Find the value of tan π/8.
Solution:
Method 1:

On cross multiplication we get

Here a = 1, b = 2, c = -1

Method 2:

Method 3:
consider tan A

Question 15.

Solution:
Given that tan 

We wish to find tan(α + 2β)