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

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2

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

ii) 2 tan2 45° + cos2 30° – sin260°
= 2(tan 45°)2 + (cos 30°)2 – (sin 60°)2
= 2 (1)2 + \(\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= 2 × 1
= 2
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 3
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 4
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 5

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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°

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

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

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\)
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 Q2
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 7

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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.
Solution:
tan (A + B) = \(\sqrt{3}\)
tan (A + B) = tan 60°
A + B = 60°
tan (A – B) = \(\frac{1}{\sqrt{3}}\) = tan 30°
tan(A – B) = tan 30°
A – B = 30° → (2)
Adding (1) and (2)
A + B + A – B = 60 + 30
2A = 90
A = \(\frac{90}{2}\) = 45°
A = 45°
Put A = 45° in eqn (1)
A + B = 60
B = 60 – A= 60 – 45°
B = 15°.

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin6 increases as θ increases.
(iii) The value of cosθ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
(i) False:
Let us take A = 30° and B = 60°, then
L.H.S = sin (30° + 60°) = sin 90° = 1
R.H.S.= sin 30° + sin 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 Q4
∴ L.H.S. ≠ R.H.S.
(ii) True:
Since, the values of sin θ increases from 0 to 1 as θ increases from 0° to 90°.
(iii) False:
Since, the value of cos θ decreases from 1 to 0 as θ increases from 0° to 90°.
(iv) False:
Let us take θ = 30°
sin 30° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\)
⇒ sin 30° ≠ cos 30°
(v) True:
We have, cot 0° = not defined

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