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Class 10 Mathematics by VP Classes

Class 10 : Notes | EduRev

The document Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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1. Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin60

 Notes | EduRev
Solution:
(i) sin 60° cos 30° + sin 30° cos 60°
First, find the values of the given trigonometric ratios
sin 30° = 1/2
cos 30° = √3/2
sin 60° = 3/2
cos 60°= 1/2
Now, substitute the values in the given problem
sin 60° cos 30° + sin 30° cos 60° = √3/2 ×√3/2 + (1/2) ×(1/2 ) = 3/4+1/4 = 4/4 =
(ii) 2 tan2 45° + cos2 30° – sin2 60
We know that, the values of the trigonometric ratios are:
sin 60° = √3/2
cos 30° = √3/2
tan 45° = 1
Substitute the values in the given problem
2 tan2 45° + cos2 30° – sin2 60 = 2(1)+ (√3/2)2-(√3/2)2
2 tan2 45° + cos2 30° – sin2 60 = 2 + 0
2 tan2 45° + cos2 30° – sin2 60 = 2
(iii) cos 45°/(sec 30°+cosec 30°)
We know that,

cos 45° = 1/√2

sec 30° = 2/√3

cosec 30° = 2

Substitute the values, we get
 Notes | EduRev
Now, multiply both the numerator and denominator by √2 , we get
 Notes | EduRev
Therefore, cos 45°/(sec 30°+cosec 30°) = (3√2 – √6)/8
 Notes | EduRev
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
 Notes | EduRev
 Notes | EduRev
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°)/(sin30° + cos30°)
= 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) (A) is correct.
Substitute the of tan 30° in the given equation
tan 30° = 1/√3
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°
The obtained solution is equivalent to the trigonometric ratio sin 60°
(ii) (D) is correct.
Substitute the of tan 45° in the given equation
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.
To find the value of A, substitute the degree given in the options one by one
sin 2A = 2 sin A is true when A = 0°
As sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 2 × 0 = 0
or,
Apply the sin 2A formula, to find the degree value
sin 2A = 2sin A cos A
⇒ 2sin A cos A = 2 sin A
⇒ 2cos A = 2 ⇒ cos A = 1
Now, we have to check, to get the solution as 1, which degree value has to be applied.
When 0 degree is applied to cos value, i.e., cos 0 = 1
Therefore, ⇒ A = 0°
(iv) (C) is correct.
Substitute the of tan 30° in the given equation
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°
The value of the given equation is equivalent to 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
Since √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
Since 1/√3 = tan 30°
Now substitute the degree value
⇒ tan (A – B) = tan 30°
(A – B) = 30° … equation (ii)
Now add the equation (i) and (ii), we get
A + B + A – B = 60° + 30°
Cancel the terms B
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° 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.
Justification:
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.
Justification:
According to the values obtained as per the unit circle, the values of sin are:
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. Hence, the statement is true
(iii) False.
According to the values obtained as per the unit circle, the values of cos are:
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) False
sin θ = cos θ, when a right triangle has 2 angles of (π/4). Therefore, the above statement is false.
(v) True.
Since cot function is the reciprocal of the tan function, it is also written as:
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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