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

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

Trigonometry Exercise 11.2 Solutions Question 1.
Evaluate the following :
i) sin 60° cos 30° + sin 30° cos 60°
ii) 2 tan2 45° + cos2 30° – sin2 60° Solution:
i) sin 60° cos 30° + sin 30° cos 60° ii) 2 tan2 45° + cos2 30° – sin260°
= 2(tan 45°)2 + (cos 30°)2 – (sin 60°)2
= 2 (1)2 + (3√2)2−(3√2)2
= 2 × 1
= 2   Exercise 11.2 Class 10 Trigonometry Question 2.
Choose the correct option and justify your choice :
i) 2tan30∘1+tan230∘ =
A) sin 60°
B) cos 60°
C) tan 60°
D) sin 30°
Solution:
A) sin 60° ii) 1−tan245∘1+tan245∘ =
A) tan 90°
B) 1
C) sin 45°
D) 0
Solution:
D) 0 iii) sin 2A = 2 sin A is true when A ;
A) 0°
B) 30°
C) 45°
D) 60°
Solution:
A) 0°
LHS = sin 2A = sin 0° = 0
RHS = 2sin A = 2.sin0° = 0

iv) 2tan30∘1−tan230∘ =
A) cos 60°
B) sin 60°
C) tan 60°
D) sin 30°
Solution:
C) tan 60°  Introduction To Trigonometry Exercise 11.2 Question 3.
If tan (A + B) =3–√ and tan (A – B) = 13√ 0° < A + B ≤ 90°; A > B. find A and B.
Solution:
tan (A + B) = 3–√
tan (A + B) = tan 60°
A + B = 60°
tan (A – B) = 13√ = tan 30°
tan(A – B) = tan 30°
A – B = 30° → (2)
A + B + A – B = 60 + 30
2A = 90
A = 902 = 45°
A = 45°
Put A = 45° in eqn (1)
A + B = 60
B = 60 – A= 60 – 45°
B = 15°.

Introduction To Trigonometry Class 10 Exercise 11.2 Question 4.
i) sin (A + B) = sin A + sin B
Solution:
False
Take A = 45° and B = 45°
LHS: sin (45° + 45°) = sin 90° = 1
RHS: sin 45 + sin 45 = 12√+12√=22√
LHS ≠ RHS.

ii) The value of sin θ increases as θ increases.
Solution:
True. ∴ The value of sin θ increases as θ increases.

iii) The value of cos θ increases as θ increases.
Solution:
False.
cos 0° = 1 cos 30° = 3√2 =0.87
cos 45° = 12√ = 0.7 cos 60° = 12= = 0.5
cos 90° = 1
∴ The value of cos 0 increases as 0 increases – the statement is False.

iv) sin θ = cos θ for all values of θ.
Solution:
False.
sin 30° = 12 cos 30° = 3√2
sin 30° ≠ cos 30°
But sin 45° = cos 45° = 12√

v) cot A is not defined for A = 0°.
Solution:
True. ∴ When A = 0°, cot A is not defined.

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