NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2

NCERT Solutions Class 10 Maths
Chapter – 8 (Introduction to Trigonometry)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 8 Introduction to Trigonometry Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.

Chapter : 8 Introduction to Trigonometry

Exercise – 8.2

1. Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan² 45° + cos² 30° – sin² 60
(iii) cos 45°/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5 cos²60° + 4 sec²30° – tan²45°)/(sin²30° + cos²30°)

Solution –

(i) sin 60° cos 30° + sin 30° cos 60°
Using the values of the given trigonometric ratios
sin 60° cos 30° + sin 30° cos 60°
= √3/2 × √3/2 + (1/2) × (1/2 )
= 3/4 + 1/4
= 4/4
=1

(ii) 2 tan² 45° + cos² 30° – sin² 60
Using the values of the given trigonometric ratios
2 tan² 45° + cos² 30° – sin² 60
= 2(1)²+ (√3/2)²– (√3/2)²= 2 + 0
= 2

(iii) cos 45°/ (sec 30°+cosec 30°)
Using the values of the given trigonometric ratios
= $\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$ = $\frac{\frac{1}{\sqrt{2}}}{\frac{2 +2\sqrt{3}}{\sqrt{3}}}$

= $\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 +2\sqrt{3}}$

= $\frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)}$
Now, rationalize the terms we get

= $\frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)}$ × $\frac{\sqrt{3} -1}{\sqrt{3} -1}$

= $\frac{3-\sqrt{3}}{4\sqrt{2}}$
Now, multiply both the numerator and denominator by √2 , we get

= $\frac{3-\sqrt{3}}{4\sqrt{2}}$ × $\frac{\sqrt{2}}{\sqrt{2}}$

= $\frac{3\sqrt{2}-\sqrt{6}}{8}$

(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
Using the values of the given trigonometric ratios

= $\frac{\frac{1}{2} + 1 -\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2} +1}$ = $\frac{\frac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\frac{4+\sqrt{3+2\sqrt{3}}}{2\sqrt{3}}}$

= $\frac{3\sqrt{3}-4}{3\sqrt{3}+4}$

Now, rationalize the terms we get

= $\frac{3\sqrt{3}-4}{3\sqrt{3}+4}$ × $\frac{3\sqrt{3}-4}{3\sqrt{3}-4}$

= $\frac{27-12\sqrt{3}-12\sqrt{3}+16}{27-12\sqrt{3} +12\sqrt{3}+16}$

= $\frac{27-14\sqrt{3}+16}{11}$

= $\frac{43-24\sqrt{3}}{11}$

(v) (5cos²60° + 4sec²30° – tan²45°)/(sin²30° + cos²30°)
Using the values of the given trigonometric ratios
= 5(1/2)²+ 4(2/√3)²– 1²/(1/2)²+ (√3/2)²= (5/4 + 16/3 – 1)/(1/4 + 3/4)
= (15 + 64 – 12)/12/(4/4)
= 67/12

2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan²30° =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) 1-tan²45°/1+tan²45° =
(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) 2tan30°/1-tan²30° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

Solution –

(i) 2tan 30°/1+tan²30° =
= 2(1/√3)/1 + (1/√3)²= (2/√3)/(1 + 1/3)
= (2/√3)/(4/3)
= 6/4√3
= √3/2
= sin 60°
(A) is correct.

(ii) 1 – tan²45°/1 + tan²45° =
= (1 – 1²)/(1 + 1²)
= 0/2
= 0
(D) is correct.

(iii) sin 2A = 2 sin A is true when A =
sin 2A = sin 0° = 0
2 sin A = 2 sin 0°
= 2 × 0
= 0
sin 2A = 2sin A cos A
⇒2sin A cos A = 2 sin A
⇒ 2cos A = 2
⇒ cos A = 1
Therefore,
⇒ A = 0°
(A) is correct.

(iv) 2tan30°/1-tan²30° =
= 2(1/√3)/1 – (1/√3)²= (2/√3)/(1 – 1/3)
= (2/√3)/(2/3)
= √3
= tan 60°
(C) is correct.

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
√3 = tan 60°
Now substitute the degree value
tan (A + B) = tan 60°
(A + B) = 60° ———– (i)
The above equation is assumed as equation (i)
tan (A – B) = 1/√3
1/√3 = tan 30°
Now substitute the degree value
tan (A – B) = tan 30°
(A – B) = 30° ———– (ii)
Now 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) to find the value of B
45° + B = 60°
B = 60° – 45°
B = 15°
Therefore
A = 45°
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) sin (A + B) = sin A + sin B.
Let A = 30° and B = 60°, then
sin (A + B) = sin (30° + 60°) = sin 90° = 1
sin A + sin B = sin 30° + sin 60°
= 1/2 + √3/2
= (1 + √3)/2
False

(ii) The value of sin θ increases as θ increases.
sin 0° = 0
sin 30° = 1/2
sin 45° = 1/√2
sin 60° = √3/2
sin 90° = 1
Thus the value of sin θ increases as θ increases.
True

(iii) The value of cos θ increases as θ increases.
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 given above is False.

(iv) sin θ = cos θ for all values of θ.
sin θ = cos θ,
when a right triangle has 2 angles of (π/4).
Therefore, the above statement is False.

(v) cot A is not defined for A = 0°.
cot A = cos A/sin A
Now substitute A = 0°
cot 0° = cos 0°/sin 0° = 1/0 = undefined.
Hence, it is True

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NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2