**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 tan^{2} 45° + cos2 30° – sin^{2} 60°

Solution:

i) sin 60° cos 30° + sin 30° cos 60°

ii) 2 tan^{2} 45° + cos2 30° – sin^{2}60°

= 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

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:

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°

∴ 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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