NCERT Solutions class 10 maths chapter 8

Introduction to Trigonometry

NCERT Solutions class 10 maths chapter 8 Exercise 8.2

Question 1

Evaluate the following:

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan² 45° + cos² 30° – sin² 60

Answer

(i) sin 60° cos 30° + sin 30° cos 60°

sin 30° = 1/2

cos 30° = √3/2

sin 60° = 3/2

cos 60°= 1/2

on putting value

sin 60° cos 30° + sin 30° cos 60°
= × + × = + = 4/4 =

(ii) 2 tan² 45° + cos² 30° – sin² 60

We know that

sin 60° = √3/2

cos 30° = √3/2

tan 45° = 1

On putting value

2 tan² 45° + cos² 30° – sin² 60
= 2(1)² + (√3/2)²-(√3/2)²

= 2 + 0

= 2

(iii)

We know that,

cos 45° = 1/√2

sec 30° = 2/√3

cosec 30° = 2

On putting value we get

We know that,

sin 30° = 1/2

tan 45° = 1

cosec 60° = 2/√3

sec 30° = 2/√3

cos 60° = 1/2

cot 45° = 1

Substitute the values in the given problem, we get

We know that,

cos 60° = 1/2

sec 30° = 2/√3

tan 45° = 1

sin 30° = 1/2

cos 30° = √3/2

Now, substitute the values in the given problem, we get

(5cos²60° + 4sec²30° – tan²45°)/(sin² 30° + cos² 30°)

Question 1

Choose the correct option and justify your choice :
(i)
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) =
(A) tan 90° (B) 1 (C) sin 45° (D) 0

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

(iv)

=
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

Answer

(i) (A) is correct.

tan 30° = 1/√3


= + (1/√3)²

= =

= 6/4√3 = √3/2 = sin 60°

(ii) (D) is correct.

tan 45° = 1

1-tan²45°/1+tan²45°
= (1-1²)/(1+1²)

= 0/2 = 0

The solution of the above equation is 0.

(iii) (A) is correct.

As sin 2A = 2sin A cos A

=2sin A cos A = 2 sin A

= 2cos A = 2 = cos A = 1

here cos 0 =1

Therefore, A = 0°

(iv) (C) is correct.

tan 30° = 1/√3

2tan30°/1-tan²30° = 2(1/√3)/1-(1/√3)²

= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°

3. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.

Solution:

tan (A + B) = √3

here √3 = tan 60°

On compare

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

Similarly

tan (A – B) = 1/√3

Here 1/√3 = tan 30°

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

2A = 90°

A= 45°

Now, substitute the value of A in equation (i)

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

4. State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ 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.

Justify:

Let us take A = 30° and B = 60°, then

Substitute the values in the sin (A + B) formula, we get

sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,

sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = 1+√3/2

Since the values obtained are not equal, the solution is false.

(ii) True.

Justify:

As we know

sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2

sin 90° = 1

As we saw on θ increases sin θ is also increases as. So the statement is true

(iii) False.

As we know
cos 0° = 1, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, cos 90° = 0

Thus, the value of cos θ decreases as θ increases. So, the statement is false.

(iv) False

sin θ = cos θ,let θ=30°
So sin 30° = 1/2, cos 30° = √3/2, that are not equal

(v) True.

cot A = cos A/sin A

here A = 0°

cot 0° = cos 0°/sin 0° = 1/0 = undefined.

Hence, it is true