Samacheer Kalvi 11th Business Maths Guide Chapter 4 Trigonometry Ex 4.2

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.2

Samacheer Kalvi 11th Business Maths Trigonometry Ex 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:
(i) cosec 15° = \(\frac{1}{\sin 15^{\circ}}\)
Consider sin 15° = sin(45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°

cosec 15° = \(\frac{1}{\sin 15^{\circ}}\) = \(\frac{2 \sqrt{2}}{\sqrt{3}-1}\)

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

(iii) cot 75° = \(\frac{1}{\tan 75^{\circ}}\)
Consider tan 75° = tan (30° + 45°)

cot 75° = \(\frac{1}{\tan 75^{\circ}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\)

Question 2.
Find the values of the following:
(i) sin 76° cos 16° – cos 76° sin 16°
(ii) \(\sin \frac{\pi}{4} \cos \frac{\pi}{12}+\cos \frac{\pi}{4} \sin \frac{\pi}{12}\)
(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°
= \(\frac{\sqrt{3}}{2}\)

(ii) This is of the form sin(A + B) = \(\sin \left(\frac{\pi}{4}+\frac{\pi}{12}\right)\)
= \(\sin \left(\frac{3 \pi+\pi}{12}\right)\)
= \(\sin \frac{4 \pi}{12}\)
= \(\sin \frac{\pi}{3}\)
= \(\frac{\sqrt{3}}{2}\) (∵ sin 60° = \(\frac{\sqrt{3}}{2}\))

(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°
= \(\frac{\sqrt{3}}{2}\)

Question 3.
If sin A = \(\frac{3}{5}\), 0 < A < \(\frac{\pi}{2}\) and cos B = \(\frac{-12}{13}\), π < B < \(\frac{3 \pi}{2}\), find the values of the following:
(i) cos(A + B)
(ii) sin(A – B)
(iii) tan(A – B)
Solution:
Given that sin A = \(\frac{3}{5}\), 0 < A < \(\frac{\pi}{2}\) (i.e., A lies in first quadrant)
Since A lies in first quadrant cos A is positive.

cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}=\frac{4}{5}\)
tan A = \(\frac{3}{4}\)
AB = \(\sqrt{5^{2}-3^{2}}\) = 4
Also given that cos B = \(\frac{-12}{13}\), π < B < \(\frac{3 \pi}{2}\) (i.e., B lies in third quadrant)
Now sin B lies in third quadrant. sin B is negative.

CA = \(\sqrt{13^{2}-12^{2}}\) = 5
sin B = \(\frac{-\text { Opposite side }}{\text { Hypotenuse }}=\frac{-5}{13}\)
tan B = \(\frac{-\text { Opposite side }}{\text { Adjacent }}=\frac{5}{12}\) [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.
If cos A = \(\frac{13}{14}\) and cos B = \(\frac{1}{7}\) where A, B are acute angles prove that A – B = \(\frac{\pi}{3}\)
Solution:
cos A = \(\frac{13}{14}\), cos B = \(\frac{1}{7}\)

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

cos(A – B) = cos 60°
A – B = 60° = \(\frac{\pi}{3}\)

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.
If cot α = \(\frac{1}{2}\), sec β = \(\frac{-5}{3}\), where π < α < \(\frac{3 \pi}{2}\) and \(\frac{\pi}{2}\) < β < π, find the value of tan(α + β). State the quadrant in which α + β terminates.
Solution:
Given that cot α = \(\frac{1}{2}\) where π < α < \(\frac{3 \pi}{2}\) (i.e,. α lies in third quadrant)
tan α = \(\frac{1}{\frac{1}{2}}\) = 2 [∵ In 3rd quadrant tan α is positive]
Also given that sec β = \(\frac{-5}{3}\) where \(\frac{\pi}{2}\) < β < π (i.e., β lies in second quadrant cos β and tan β are negative)


tan (α + β) = \(\frac{2}{11}\) which is positive.
α + β terminates in first quandrant.

Question 7.
If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan 22\(\frac{1}{2}\).
Solution:
Given A + B = 45°
tan (A + B) = tan 45°
\(\frac{\tan A+\tan B}{1-\tan A \tan B}=1\)
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)
Put A = B = 22\(\frac{1}{2}\) in (1) we get
(1 + tan 22\(\frac{1}{2}\)) (1 + tan 22\(\frac{1}{2}\)) = 2
⇒ (1 + tan22\(\frac{1}{2}\))² = 2
⇒ 1 + tan 22\(\frac{1}{2}\) = ±√2
⇒ tan 22\(\frac{1}{2}\) = ±√2 – 1
Since 22\(\frac{1}{2}\) is acute, tan 22\(\frac{1}{2}\) is positive and therefore tan 22\(\frac{1}{2}\) = √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°
= 2 sin A \(\left(\frac{1}{2}\right)\)
= sin A
= RHS

(ii) 4A = 3A + A
tan 4A = tan (3A + A)
tan 4A = \(\frac{\tan 3 \mathrm{A}+\tan \mathrm{A}}{1-\tan 3 \mathrm{A} \tan \mathrm{A}}\)
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 = \(\frac{12}{13}\), find sin 3A.
Solution:
(i) tan θ = 3

(ii) If sin A = \(\frac{12}{13}\)
We know that sin 3A = 3 sin A – 4 sin³ A

Question 10.
If sin A = \(\frac{3}{5}\), find the values of cos 3A and tan 3A.
Solution:

Given sin A = \(\frac{3}{5}\)
cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}=\frac{4}{5}\)
and tan A = \(\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{3}{4}\)
We know that cos 3A = 4 cos³ A – 3 cos A

Question 11.
Prove that \(\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0\)
Solution:
Consider \(\frac{\sin (B-C)}{\cos B \cos C}\)
= \(\frac{\sin \mathrm{B} \cos \mathrm{C}-\cos \mathrm{B} \sin \mathrm{C}}{\cos \mathrm{B} \cos \mathrm{C}}\)
= \(\frac{\sin B \cos C}{\cos B \cos C}-\frac{\cos B \sin C}{\cos B \cos C}\)
= tan B – tan C ……… (1)
Similarly we can prove \(\frac{\sin (C-A)}{\cos C \cos A}\) = tan C – tan A …….(2)
and \(\frac{\sin (A-B)}{\cos A \cos B}\) = tan A – tan B …….. (3)
Add (1), (2) and (3) we get
\(\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0\)

Question 12.
If tan A – tan B = x and cot B – cot A = y prove that cot(A – B) = \(\frac{1}{x}+\frac{1}{y}\).
Solution:


Hence proved.

Question 13.
If sin α + sin β = a and cos α + cos β = b, then prove that cos(α – β) = \(\frac{a^{2}+b^{2}-2}{2}\)
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)
∴ cos(α – β) = \(\frac{a^{2}+b^{2}-2}{2}\)

Question 14.
Find the value of tan\(\frac{\pi}{8}\).
Solution:
Method 1:
\(\frac{\pi}{8}=\frac{180^{\circ}}{8}=\frac{45^{\circ}}{2}=22 \frac{1}{2}\)
We know that tan 2A = \(\frac{2 \tan A}{1-\tan ^{2} A}\)
Put A = 22\(\frac{1}{2}\) in the above formula

On cross multiplication we get


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

Since 22\(\frac{1}{2}\) is acute tan 22\(\frac{1}{2}\) is positive tan 22\(\frac{1}{2}\) = tan \(\frac{\pi}{8}\)
= -1 + √2
= √2 – 1

Method 2:



∴ \(\tan ^{2} 22 \frac{1}{2}=(\sqrt{2}-1)^{2}\)
Taking square root, \(\tan ^{2} 22 \frac{1}{2}\) = ±(√2 – 1)
But \(22 \frac{1}{2}\) lies in first quadrant, tan \(22 \frac{1}{2}\) is positive.
∴ tan 22\(\frac{1}{2}\) = √2 – 1

Method 3:
consider tan A = \(\frac{\sin 2 A}{1+\cos 2 A}\)
Put A = \(22 \frac{1}{2}\)



tan 22\(\frac{1}{2}\) = √2 – 1

Question 15.
If tan α = \(\frac{1}{7}\), sin β = \(\frac{1}{\sqrt{10}}\). Prove that α + 2β = \(\frac{\pi}{4}\) where 0 < α < \(\frac{\pi}{2}\) and 0 < β < \(\frac{\pi}{2}\).
Solution:
Given that tan α = \(\frac{1}{7}\)
We wish to find tan(α + 2β)