The cumulative frequency with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than n/2 (i.e. 100/2 = 50) is 76, belonging to class interval 7 – 10. Median class = 7-10 lower limit (l) of median class = 7 Cumulative frequency (cf) of clasRead more
The cumulative frequency with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than n/2 (i.e. 100/2 = 50) is 76, belonging to class interval 7 – 10.
Median class = 7-10
lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
Median = l + ((n/2 – cf)/f) × h = 7 + ((50 – 36)/40) × 3 = 7 + 42/40 = 7 + 1.05 = 8.05
To find the class marks of the given class intervals, the following relation is used.
xᵢ = (Upper limit + Lower limit)/ 2
Taking 11.5 as assumed mean (a), dᵢ, uᵢ, and fᵢuᵢ are calculated according to step deviation method as follows.
Number of letters 1-4 4-7 7-10 10-13 13- 16 16-19 Total
No. of surnames (fᵢ) 6 30 40 16 4 4 100
xᵢ 2.5 5.5 8.5 11.5 14.5 17.5
dᵢ = xᵢ – 11.5 -9 – 6 -3 0 3 6
uᵢ = dᵢ / 3 -3 2 1 0 1 2
fᵢuᵢ -18 -60 -40 0 4 8 -106
From the table, we obtain
∑fᵢ = 100, ∑fᵢuᵢ = -106, a = 11.5 and h = 3
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 11.5 + (-106/100) × 3 = 11.5 – 318/100 = 11.5 – 3.18 = 8.32
The data in the given table can be written as
Number of Letters 1-4 4-7 7-10 10-13 13-16 16-19 Total (n)
Frequency (fᵢ) 6 30 40 16 4 4 100
From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7 – 10.
Modal class = 7 – 10
Lower limit (l) of modal class = 7
Class size (h) = 3
Frequency (f₁) of modal class = 40
Frequency (f₀) of class preceding the modal class = 30
Frequency (f₂) of class succeeding the modal class = 16
Mode = l + ((f₁ – f₀)/2f₁ – f₀ – f₂) × h = 7 + ((40 – 30) / 2 × 40 – 30 -16) × 3 = 7 + 10/34 × 3 = 7 + 0.88 = 7.88
Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.
The cumulative frequency with their respective class intervals are as follows. Cumulative frequency just greater than n/2 (i.e. 30/2 = 15) is 19, belonging to class interval 55 - 60. Median class = 55 - 60 Lower limit (l) of median class = 55 Frequency (f) of median class = 6 Cumulative frequency (cRead more
The cumulative frequency with their respective class intervals are as follows.
Cumulative frequency just greater than n/2 (i.e. 30/2 = 15) is 19, belonging to class interval 55 – 60.
Median class = 55 – 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
Median = l + ((n/2 – cf)/f) × h = 55 + ((15 – 13)/6) × 5 = 55 + 10/6 = 55 + 1.67 = 56.67.
Therefore, median weight is 56.67 kg.
The cumulative frequency for the given data is calculated as follows. From the table, it can be observed that n = 60 45 + x + y = 60 or x + y = 15 ...(1) Median of thr data is given as 28.5 which lies in interval 20 - 30. Therefore, median class = 20 - 30 Lower limit (l) of median class = 20 CumulatRead more
The cumulative frequency for the given data is calculated as follows.
From the table, it can be observed that n = 60
45 + x + y = 60 or x + y = 15 …(1)
Median of thr data is given as 28.5 which lies in interval 20 – 30.
Therefore, median class = 20 – 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
Median = l + ((n/2 – cf)/(f)) × h
⇒ 28.5 = 20 + {(60/2 – (5 + x))/20} × 10
⇒ 8.5 = (25 – x)/2
⇒ 17 = 25 – x
⇒ x = 8
From equation (1), 8 + y = 15 ⇒ y = 7
Hence, the value of x and y are 8 and 7 respectively.
Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. ThereforRead more
Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.
From the table, it can be observed that n = 100
Cumulative frequency (cf) just greater than n/2 (i.e. 100/2 = 50) is 78, belonging to interval 35 – 40.
Therefore, median class = 35 – 40
Lower limit (l) of median class = 35
Frequency (f) of median class = 33
Cumulative frequency (cf) of class = 45
Frequency (f) of median class = 33
Median = l + ((n/2 – cf)/(f)) × h = 35 + {(50 – 45)/20} × 5 = 35 + 25/33 = 35 + 0.76 = 35.76
Therefore, median age is 35.76 years.
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, 1/2 = 0.5 has to be added and subtracted to upper class limits and lower-class limits respectively. Continuous class intervals with respective cumulative frequRead more
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, 1/2 = 0.5 has to be added and subtracted to upper class limits and lower-class limits respectively.
Continuous class intervals with respective cumulative frequencies he represented follows.
From the table, it can be observed that the cumulative frequency just greater than n/2 (i.e. 40/2 = 20) is 29, belonging to class interval 144.5 – 153.5.
Median class = 144.5 – 153.5
Lower Limit (l) of median class = 144.5 and class size (h) = 9
Frequency (f) of median class 12
Cumulative frequency (cf) of class preceding median class = 17
Median = l + (n/2 – cf)/f) × h = 144.5 + ((20 – 17)/12) × 9 = 144.5 + 2.25 = 146.75
Therefore, median length of leaves is 146.75mm.
See this explanation video for better understanding✌😃
The cumulative frequencies with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 - 3500. Median class = 3000 - 3500 Lower limt (l) of median class = 3000 Frequency (f) ofRead more
The cumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 – 3500.
Median class = 3000 – 3500
Lower limt (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size = 500
Median = l + ((n/2 – cf)/f) × h = 3000 + ((200 – 130)/86) × 500 = 3000 + 35000/86 = 3000 + 406.978 + 3406.978
Therefore, median lifetime of lamps is 3406.98 hours.
To find the class marks, the following relation is used. xᵢ = (Upper limit + Lower limit)/2 Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows. From the table, we obtain. ∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20 mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 1Read more
To find the class marks, the following relation is used.
xᵢ = (Upper limit + Lower limit)/2
Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows.
From the table, we obtain.
∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 135 + (7/68) × 20 = 135 + 2.058 = 137.058
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 – 145.
Modal class 125 – 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 13
Frequency (f₂) of class succeeding the modal class = 14
Mode = l + (f₁ – f₀ / 2f₁ – f₀ – f₂)h = 125 × (20 – 13 / 2 × 20 – 13 – 14) × 20 = 125 + 7/13 × 20 = 125 + 10.76 = 135.76
To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units) Number of consumers Cumulative frequency
65 – 85 4 4
85 – 105 5 4 + 5 = 9
105 – 125 13 9 + 13 = 22
125 – 145 20 22 + 20 = 42
145 – 165 14 42 + 14 = 56
165 – 185 8 56 + 8 = 64
185 – 205 4 64 + 4 = 68
From the table, we obtain n = 18
Cumulative frequency (cf) just greater than n/2 (i.e, 68/2 = 34) is 42, belonging to interval 125 – 145.
Therefore, median class = 125 – 145
Lower limit (l) of median class = 125 and class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Median = l + ((n/2 – cf)/f) × h = 125 + (34 – 22 / 20) × 20 = 125 + 12 = 137
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.
(i) (2 tan 30°)/(1 - tan²30°) Putting the value of each trigonometric ratios, we get (2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2 We know that sin 60° =Y hence the option (A) is correct. (ii)(1 - tan² 45°)/(1 + tan² 45°) Putting the value of each trigonometric ratios, wRead more
(i) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2
We know that sin 60° =Y hence the option (A) is correct.
(ii)(1 – tan² 45°)/(1 + tan² 45°)
Putting the value of each trigonometric ratios, we get
(1-(1)²)(1+(1)²) = (1-1)/(1+1) = 0/2 = 0
Hence, the option (D) is correct.
(iii) sin 2A = 2 sin A
We know that sin 0 = 0, hence, the option (A) is correct.
(iv) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1-(1/√3)²) = (1/√3)/(1 – 1/3) = (2/√3)/(2/3) = 3/√3 = √3
We know that tan 60° = √3, hence, the option (C) is correct.
Here is the video explanation of the above question😁🤗
(i) tan 48° tan 23° tan 42° tan 67° = 1 LHS = tan 48° tan 23° tan 42° tan 67° = tan 48° tan 23° cot(90°-42°) cot(90°-67°) [∵ cot(90°-θ) = tanθ] = tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° × 1/tan 48° × 1/tan 23° = 1 = RHS (ii) cos 38° cos 52° - sin 38° sin 52° = 0 LHS = cos 38° cos 52° - sinRead more
(i) tan 48° tan 23° tan 42° tan 67° = 1
LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° cot(90°-42°) cot(90°-67°) [∵ cot(90°-θ) = tanθ]
= tan 48° tan 23° cot 48° cot 23°
= tan 48° tan 23° × 1/tan 48° × 1/tan 23°
= 1 = RHS
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos 52° – cos (90°-38°) cos (90°-52) [∵ cos (90°-0°) = sinθ]
= cos 38° cos 52° – cos52° – cos 38°
= 0 = RHS
(3/2)x + (5/3)y = 7 9x - 10y = 14 Here, a₁/a₂ = (3/2)/9 = 1/6 and b₁/b₂ = (5/3)/-10 = 1/6 ⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent. For more details click here => https://www.tiwariacademy.com/ncert-solutions/class-10/maths/ See here for video Solution 👀👇
(3/2)x + (5/3)y = 7
9x – 10y = 14
Here, a₁/a₂ = (3/2)/9 = 1/6 and b₁/b₂ = (5/3)/-10 = 1/6
⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
The cumulative frequency with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than n/2 (i.e. 100/2 = 50) is 76, belonging to class interval 7 – 10. Median class = 7-10 lower limit (l) of median class = 7 Cumulative frequency (cf) of clasRead more
The cumulative frequency with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than n/2 (i.e. 100/2 = 50) is 76, belonging to class interval 7 – 10.
See lessMedian class = 7-10
lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
Median = l + ((n/2 – cf)/f) × h = 7 + ((50 – 36)/40) × 3 = 7 + 42/40 = 7 + 1.05 = 8.05
To find the class marks of the given class intervals, the following relation is used.
xᵢ = (Upper limit + Lower limit)/ 2
Taking 11.5 as assumed mean (a), dᵢ, uᵢ, and fᵢuᵢ are calculated according to step deviation method as follows.
Number of letters 1-4 4-7 7-10 10-13 13- 16 16-19 Total
No. of surnames (fᵢ) 6 30 40 16 4 4 100
xᵢ 2.5 5.5 8.5 11.5 14.5 17.5
dᵢ = xᵢ – 11.5 -9 – 6 -3 0 3 6
uᵢ = dᵢ / 3 -3 2 1 0 1 2
fᵢuᵢ -18 -60 -40 0 4 8 -106
From the table, we obtain
∑fᵢ = 100, ∑fᵢuᵢ = -106, a = 11.5 and h = 3
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 11.5 + (-106/100) × 3 = 11.5 – 318/100 = 11.5 – 3.18 = 8.32
The data in the given table can be written as
Number of Letters 1-4 4-7 7-10 10-13 13-16 16-19 Total (n)
Frequency (fᵢ) 6 30 40 16 4 4 100
From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7 – 10.
Modal class = 7 – 10
Lower limit (l) of modal class = 7
Class size (h) = 3
Frequency (f₁) of modal class = 40
Frequency (f₀) of class preceding the modal class = 30
Frequency (f₂) of class succeeding the modal class = 16
Mode = l + ((f₁ – f₀)/2f₁ – f₀ – f₂) × h = 7 + ((40 – 30) / 2 × 40 – 30 -16) × 3 = 7 + 10/34 × 3 = 7 + 0.88 = 7.88
Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.
The distribution below gives the weights of 30 students of a class.
The cumulative frequency with their respective class intervals are as follows. Cumulative frequency just greater than n/2 (i.e. 30/2 = 15) is 19, belonging to class interval 55 - 60. Median class = 55 - 60 Lower limit (l) of median class = 55 Frequency (f) of median class = 6 Cumulative frequency (cRead more
The cumulative frequency with their respective class intervals are as follows.
Cumulative frequency just greater than n/2 (i.e. 30/2 = 15) is 19, belonging to class interval 55 – 60.
See lessMedian class = 55 – 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
Median = l + ((n/2 – cf)/f) × h = 55 + ((15 – 13)/6) × 5 = 55 + 10/6 = 55 + 1.67 = 56.67.
Therefore, median weight is 56.67 kg.
If the median of the distribution given below is 28.5, find the values of x and y.
The cumulative frequency for the given data is calculated as follows. From the table, it can be observed that n = 60 45 + x + y = 60 or x + y = 15 ...(1) Median of thr data is given as 28.5 which lies in interval 20 - 30. Therefore, median class = 20 - 30 Lower limit (l) of median class = 20 CumulatRead more
The cumulative frequency for the given data is calculated as follows.
From the table, it can be observed that n = 60
See less45 + x + y = 60 or x + y = 15 …(1)
Median of thr data is given as 28.5 which lies in interval 20 – 30.
Therefore, median class = 20 – 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
Median = l + ((n/2 – cf)/(f)) × h
⇒ 28.5 = 20 + {(60/2 – (5 + x))/20} × 10
⇒ 8.5 = (25 – x)/2
⇒ 17 = 25 – x
⇒ x = 8
From equation (1), 8 + y = 15 ⇒ y = 7
Hence, the value of x and y are 8 and 7 respectively.
A life insurance agent found the following data for distribution of ages of 100 policy holders.
Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. ThereforRead more
Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.
From the table, it can be observed that n = 100
Cumulative frequency (cf) just greater than n/2 (i.e. 100/2 = 50) is 78, belonging to interval 35 – 40.
Therefore, median class = 35 – 40
Lower limit (l) of median class = 35
Frequency (f) of median class = 33
Cumulative frequency (cf) of class = 45
Frequency (f) of median class = 33
Median = l + ((n/2 – cf)/(f)) × h = 35 + {(50 – 45)/20} × 5 = 35 + 25/33 = 35 + 0.76 = 35.76
Therefore, median age is 35.76 years.
See here for Video explanation 👇😀
See lessThe lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, 1/2 = 0.5 has to be added and subtracted to upper class limits and lower-class limits respectively. Continuous class intervals with respective cumulative frequRead more
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, 1/2 = 0.5 has to be added and subtracted to upper class limits and lower-class limits respectively.
Continuous class intervals with respective cumulative frequencies he represented follows.
From the table, it can be observed that the cumulative frequency just greater than n/2 (i.e. 40/2 = 20) is 29, belonging to class interval 144.5 – 153.5.
Median class = 144.5 – 153.5
Lower Limit (l) of median class = 144.5 and class size (h) = 9
Frequency (f) of median class 12
Cumulative frequency (cf) of class preceding median class = 17
Median = l + (n/2 – cf)/f) × h = 144.5 + ((20 – 17)/12) × 9 = 144.5 + 2.25 = 146.75
Therefore, median length of leaves is 146.75mm.
See this explanation video for better understanding✌😃
See lessThe following table gives the distribution of the life time of 400 neon lamps:
The cumulative frequencies with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 - 3500. Median class = 3000 - 3500 Lower limt (l) of median class = 3000 Frequency (f) ofRead more
The cumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 – 3500.
See lessMedian class = 3000 – 3500
Lower limt (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size = 500
Median = l + ((n/2 – cf)/f) × h = 3000 + ((200 – 130)/86) × 500 = 3000 + 35000/86 = 3000 + 406.978 + 3406.978
Therefore, median lifetime of lamps is 3406.98 hours.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality.
To find the class marks, the following relation is used. xᵢ = (Upper limit + Lower limit)/2 Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows. From the table, we obtain. ∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20 mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 1Read more
To find the class marks, the following relation is used.
xᵢ = (Upper limit + Lower limit)/2
Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows.
From the table, we obtain.
See less∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 135 + (7/68) × 20 = 135 + 2.058 = 137.058
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 – 145.
Modal class 125 – 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 13
Frequency (f₂) of class succeeding the modal class = 14
Mode = l + (f₁ – f₀ / 2f₁ – f₀ – f₂)h = 125 × (20 – 13 / 2 × 20 – 13 – 14) × 20 = 125 + 7/13 × 20 = 125 + 10.76 = 135.76
To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units) Number of consumers Cumulative frequency
65 – 85 4 4
85 – 105 5 4 + 5 = 9
105 – 125 13 9 + 13 = 22
125 – 145 20 22 + 20 = 42
145 – 165 14 42 + 14 = 56
165 – 185 8 56 + 8 = 64
185 – 205 4 64 + 4 = 68
From the table, we obtain n = 18
Cumulative frequency (cf) just greater than n/2 (i.e, 68/2 = 34) is 42, belonging to interval 125 – 145.
Therefore, median class = 125 – 145
Lower limit (l) of median class = 125 and class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Median = l + ((n/2 – cf)/f) × h = 125 + (34 – 22 / 20) × 20 = 125 + 12 = 137
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.
Choose the correct option and justify your choice:
(i) (2 tan 30°)/(1 - tan²30°) Putting the value of each trigonometric ratios, we get (2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2 We know that sin 60° =Y hence the option (A) is correct. (ii)(1 - tan² 45°)/(1 + tan² 45°) Putting the value of each trigonometric ratios, wRead more
(i) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2
We know that sin 60° =Y hence the option (A) is correct.
(ii)(1 – tan² 45°)/(1 + tan² 45°)
Putting the value of each trigonometric ratios, we get
(1-(1)²)(1+(1)²) = (1-1)/(1+1) = 0/2 = 0
Hence, the option (D) is correct.
(iii) sin 2A = 2 sin A
We know that sin 0 = 0, hence, the option (A) is correct.
(iv) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1-(1/√3)²) = (1/√3)/(1 – 1/3) = (2/√3)/(2/3) = 3/√3 = √3
We know that tan 60° = √3, hence, the option (C) is correct.
Here is the video explanation of the above question😁🤗
See lessShow that the following trigonometric relation are true.
(i) tan 48° tan 23° tan 42° tan 67° = 1 LHS = tan 48° tan 23° tan 42° tan 67° = tan 48° tan 23° cot(90°-42°) cot(90°-67°) [∵ cot(90°-θ) = tanθ] = tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° × 1/tan 48° × 1/tan 23° = 1 = RHS (ii) cos 38° cos 52° - sin 38° sin 52° = 0 LHS = cos 38° cos 52° - sinRead more
(i) tan 48° tan 23° tan 42° tan 67° = 1
LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° cot(90°-42°) cot(90°-67°) [∵ cot(90°-θ) = tanθ]
= tan 48° tan 23° cot 48° cot 23°
= tan 48° tan 23° × 1/tan 48° × 1/tan 23°
= 1 = RHS
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos 52° – cos (90°-38°) cos (90°-52) [∵ cos (90°-0°) = sinθ]
= cos 38° cos 52° – cos52° – cos 38°
= 0 = RHS
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See lessOn comparing the ratios, a₁/a₂, b₁/b₂ and c₁/c₂, find out whether the pair of linear equations are consistent, or inconsistent. (3/2)x + (5/3)y = 7; 9x – 10y = 14
(3/2)x + (5/3)y = 7 9x - 10y = 14 Here, a₁/a₂ = (3/2)/9 = 1/6 and b₁/b₂ = (5/3)/-10 = 1/6 ⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent. For more details click here => https://www.tiwariacademy.com/ncert-solutions/class-10/maths/ See here for video Solution 👀👇
(3/2)x + (5/3)y = 7
9x – 10y = 14
Here, a₁/a₂ = (3/2)/9 = 1/6 and b₁/b₂ = (5/3)/-10 = 1/6
⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent.
For more details click here => https://www.tiwariacademy.com/ncert-solutions/class-10/maths/
See here for video Solution 👀👇
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