AP 10th Class Maths 8th Chapter Introduction to Trigonometry Exercise 8.2 Solutions

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Well-designed AP 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.2 offers step-by-step explanations to help students understand problem-solving strategies.

Introduction to Trigonometry Class 10 Exercise 8.2 Solutions – 10th Class Maths 8.2 Exercise Solutions

Question 1.
Evaluate the following :
i) sin 60° cos 30° + sin 30° cos 60°
ii) 2 tan2 45° + cos2 30° – sin2 60°
iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+cosec 30^{\circ}}\)

iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-cosec 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)

v) \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)
Solution:
i) sin 60° cos 30° + sin 30° cos 60°
From the trigonometric ratio value table,
sin 30° = \(\frac{1}{2}\), cos 30° = \(\frac{\sqrt{3}}{2}\),
sin 60° = \(\frac{\sqrt{3}}{2}\), cos 60° = \(\frac{\sqrt{3}}{2}\)
sin 60° cos 30° + sin 30° cos 60°
= \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\) + \(\frac{1}{2} \times \frac{1}{2}\) = \(\frac{3}{4}+\frac{1}{4}\) = \(\frac{4}{4}\) = 1
∴ sin 60° cos 30° + sin 30° cos 60° = 1.

ii) 2 tan2 45° + cos2 30° – sin2 60°
From the trigonometric ratio value table, tan 45° = 1, cos 30° = \(\frac{\sqrt{3}}{2}\), sin 60° = \(\frac{\sqrt{3}}{2}\)
AP 10th Class Maths 8th Chapter Introduction to Trigonometry Exercise 8.2 Solutions 1

iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+cosec 30^{\circ}}\)
From the trigonometric ratio value
table, cos 45° = \(\frac{1}{\sqrt{2}}\), sec 30° = \(\frac{2}{\sqrt{3}}\), cosec 30° = 2
\(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+cosec 30^{\circ}}\)
AP 10th Class Maths 8th Chapter Introduction to Trigonometry Exercise 8.2 Solutions 2

iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-cosec 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
From the trigonometric ratio value table,
sin 30° = \(\frac{1}{2}\), tan 45° = 1,
cosec 60° = \(\frac{2}{\sqrt{3}}\), sec 30° = \(\frac{2}{\sqrt{3}}\)
cos 60° = \(\frac{1}{2}\), cot 45° = 1
\(\frac{\sin 30^{\circ}+\tan 45^{\circ}-cosec 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
AP 10th Class Maths 8th Chapter Introduction to Trigonometry Exercise 8.2 Solutions 3

v) \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)
From trigonometric ratio value table,
sin 30° = \(\frac{1}{2}\), cos 30° = \(\frac{\sqrt{3}}{2}\)
sec 30° = \(\frac{2}{\sqrt{3}}\), tan 45° = 1
AP 10th Class Maths 8th Chapter Introduction to Trigonometry Exercise 8.2 Solutions 4

Question 2.
Choose the correct option and justify your choice:
i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)
A) sin 60°
B) cos 60°
c) tan 60°
D) sin 30°
Solution:
A) sin 60°

From trigonometric ratio value table,
AP 10th Class Maths 8th Chapter Introduction to Trigonometry Exercise 8.2 Solutions 5

ii) \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}\) =
A) tan 90°
B) 1
C) sin 45°
D) 0
Solution:
D) 0

tan 45° = 1
\(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}\) = \(\frac{1-1^2}{1+1^2}\) = \(\frac{1-1}{1+1}\) = \(\frac{0}{2}\) = 0

iii) sin 2A = 2 sin A is true when A =
A) 0° B) 30° C) 45° D) 60°
Sol.
A) 0°

From the trigonometric ratio value table,
If A = 0° then sin A = sin 0° = 0,
2 sin A = 0 and sin 2A = sin 2(0°) = sin 0° = 0
sin 2A = 2 sin A = 0

iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)
A) cos 60°
B) sin 60°
C) tan 60°
D) sin 30°
Solution:
C) tan 60°

We know, tan 30° = 2 tan 30°
AP 10th Class Maths 8th Chapter Introduction to Trigonometry Exercise 8.2 Solutions 6

Question 3.
If tan (A + B) = 73 and tan (A – B) =
0° < A + B < 90°; A > B, find’A and B.
Sol. Given tan (A + B) = 73 and

3. If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\). 0° < A + B ≤ 90; A > B. find A and B.
Solution:
we know tan (A + B)= \(\sqrt{3}\) = tan 30°
So, A + B = 30° ……..(1)
tan (A + B)= \(\frac{1}{\sqrt{3}}\) = tan 60°
tan (A – B) = tan 60°
(1) + (2)
A + B + A – B = 60° + 30°
2A = 90° ⇒ A = 45°
Put A = 45° in (1) ⇒ 45 + B = 60°
B = 60° – 45° = 15°
Therefore A = 45° and B = 15°

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sinA + sin B.
(ii) The value of sin 0 increases as 0 increases.
(iii) The value of cos 0 increases as 0 increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
(i) False.

Let A = 60°, B = 30°
Then, sin(A + B) = sin(60°+30°)
= sin 90° = 1
sin A + sin B = sin 60° + sin 30°
= \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) = \(\frac{\sqrt{3}+1}{2}\)
So, sin (A + B) ≠ sinA + sin B.

ii) True.

sin 0° = 0, sin 30° = \(\frac{1}{2}\),
sin 45° = \(\frac{1}{\sqrt{2}}\), sin 60° = \(\frac{\sqrt{3}}{2}\) and sin 90° = 1
So, if θ increases, then the value of sin θ increases.

iii) False
cos 0° = 1, cos 30° = \(\frac{\sqrt{3}}{2}\),
cos 45° = \(\frac{1}{\sqrt{2}}\), cos 60° = \(\frac{1}{2}\) and cos 90° = 0
So, if θ increases, then the value of sin θ decreases.

iv) False.

sin 0° = 0 but cos 0° = 1
So, sin 0° ≠ cos 0°
sin 30° = \(\frac{\sqrt{3}}{2}\) but
So, sin 30° ≠ cos 30°
sin 45° = \(\frac{1}{\sqrt{2}}\) but cos 45° = \(\frac{1}{\sqrt{2}}\)
So, sin 45° ≠ cos 45°
So, sin θ = cos θ for all values.

v) True.

cot A = cot θ = \(\frac{cos 0°}{sin 0°}\) = \(\frac{1}{0}\)
= Not defined.
So, cot A is not defined for for A = 0°.

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