KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2

Students can Download Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2

1. Evaluate the following:
i) sin 60° cos 30° + sin 30° cos 60°
ii) 2 tan2 45° + cos2 30° – sin2 60°
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 1
Answer:
i) sin 60° cos 30° + sin 30° cos 60°
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 2

ii) 2 tan2 45° + cos2 30° – sin2 60°
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 3
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 4
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 5
Rationalize the denominator and simplify
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 6
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 7

KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2

Question 2.
Choose the correct option and justify your choice:
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 8
A) sin 60°
B) cos 60°
C) tan 60°
D) sin 30°
Answer:
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 9
= sin 60°
∴ correct answer is A) sin 60°.

ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
A) tan 90°
B) 1
C) sin 45°
D) 0
Answer:
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 10
∴ correct answer is D) 0

iii) sin 2A = 2 sin A is true when A =
A) 0°
B) 30°
C) 45°
D) 60°
Answer:
when A = 0°
sin2A = 2sin A
sin(0) = 2sin 0
0 = 0
∴ correct answer is 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°
Answer:
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 11
= tan 60°
∴ correct answer is C) tan 60°.

KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2

Question 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.
Answer:
tan (A + B) = \(\sqrt{3}\) ⇒ A + B = 60° ………….. (i)
tan (A – B) = \(\frac{1}{\sqrt{3}}\) ⇒ A – B = 30° ………… (ii)
From Eqn. (i) + Eqn.(ii), we have
2A = 90 ⇒ = 45°
From eqn. (i),
A + B = 60°
45° + B = 60°
∴ B = 15

Question 4.
State whether the following are true or false. Justify your answer.
i) sin (A+B) = sin A + sin B.
Answer:
False. ∵ A = 60°, B = 30°
LHS = sin (A + B) = sin (60 + 30) = sin 90° = 1
RHS = sin A + Sin B = sin 60° + sin 30°
= \(\frac{\sqrt{3}}{2}+\frac{1}{2}\) ≠ 1
∴ LHS ≠ RHS.

ii) The value of sin θ increases as θ increases.
Answer:
True
From the table of standard angles it is true.
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 12

iii) The value of cos θ increases as θ increases.
Answer:
False
From the table of standard angles, it is false, because values decrease when θ increases.
KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2 13

iv) sin θ = cos θ for all values of θ.
Answer:
False
From the values of standard angles tables, it is wrong. It is true only for 45°.

v) cot A is not defined for A = 0°.
Answer:
True
From the table of value cot 0° = N.D
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KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.2