1. It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below. Weight (in kg) Less than38 38-40 40-42 42-44 44 46 46-48 48-50 50-52 Total(n) Frequency (f) 0 3-0=3 5-3=2 9-5=4 14-9=5 28-14=14 32-28=4 35-Read more

    It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below.

    Weight (in kg) Less than38 38-40 40-42 42-44 44 46 46-48 48-50 50-52 Total(n)
    Frequency (f) 0 3-0=3 5-3=2 9-5=4 14-9=5 28-14=14 32-28=4 35-32=3 35
    Cumulative
    frequency 0 3 5 9 14 28 32 35

    The cumulative frequency just greater than n/2 (i.e. 35/2 = 17.5) is 28, belonging to class intervals 46- 48.
    Median class = 46 – 48
    lower class limit (l) of median class = 46
    Cumulative frequency (cf) of class preceding median class = 14
    Class size (h) = 2
    Median = l + ((n/2 – cf)/f) × h = 46 + ((17.5 – 14)/14) × 2 = 46 + 0.5 = 46.5
    Therefore, median of this data is 46.5.
    Hence, the value of median is verified.

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  2. There are a total of 5 days. Shyam can go to the shop in 5 ways and Ekta can go to the shop in 5 ways. Therefore, total number of outcomes = 5 × 5 = 25 (i) They can reach on the same day in 5 ways. ie, (t, t), (w, w), (th, th), (f, f),(s, s) P(both will reach on same day) = 5/25 = 1/5 (ii) They canRead more

    There are a total of 5 days. Shyam can go to the shop in 5 ways and Ekta can go to the shop in 5 ways.
    Therefore, total number of outcomes = 5 × 5 = 25
    (i) They can reach on the same day in 5 ways.
    ie, (t, t), (w, w), (th, th), (f, f),(s, s)
    P(both will reach on same day) = 5/25 = 1/5
    (ii) They can reach on consecutive days in these 8 ways- (t, w), (w, th), (th, f), (f, s), (w,t), (th, w), (f, th), (s, f).
    Therefore, P (both will reach on consecutive days) = 8/25
    P (both will reach on same day) = 1/5(From (i)]
    P (both will reach on different days) = 1 – 1/5 = 4/5

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  3. (i) (cosec θ - cot θ)² = (1 - cos θ)(1 + cos θ) LHS = (cosec θ - cot θ)² = (1/sin θ - cos θ/sin θ) = ((1 - cos θ)/sin θ)² = ((1 - cos θ)²)/(sin² θ) = ((1 - cos θ)²)(1 - cos² θ) [∵ sin² θ = 1 - cos² θ] = ((1 - cos θ)(1 - cos θ))/((1 - cos θ)(1 + cos θ)) = (1 - cos θ)/(1 + cos θ) = RHS (ii) (cos A)/(1Read more

    (i) (cosec θ – cot θ)² = (1 – cos θ)(1 + cos θ)
    LHS = (cosec θ – cot θ)²
    = (1/sin θ – cos θ/sin θ)
    = ((1 – cos θ)/sin θ)²
    = ((1 – cos θ)²)/(sin² θ)
    = ((1 – cos θ)²)(1 – cos² θ) [∵ sin² θ = 1 – cos² θ]
    = ((1 – cos θ)(1 – cos θ))/((1 – cos θ)(1 + cos θ))
    = (1 – cos θ)/(1 + cos θ) = RHS

    (ii) (cos A)/(1 + sin A) + (1 + sin A)/(cos A) = 2 sec A
    LHS = (cos A)(1 + sin A) + (1 + sin A)/(cos A)
    = (cos² A) + (1 + sin A)²/((1 + sin A)cos A)
    = (cos² A + 1 + sin² A + 2 sin A)/((1 + sin A)cos A)
    = (1 + 1 + 2 sin A)/((1 + sin A)cos A)) [∵ sin² A + cos² A = 1]
    = (2 + 2 sin A)/((1 + sin A)cos A)
    = (2(1 + sin A))((1 + sin A)cos A) = 2/cos A = 2 sec A = RHS

    (iii) tan θ /(1 – cot θ) + cot θ /(1 – tan θ) = 1 + sec θ cosec θ
    LHS = tan θ /(1 – cot θ) + cot θ /(1 – tan θ)
    = [(sin θ /cos θ)/(1 – (cos θ /sin θ))] + [(cos θ /sin θ)/(1 – sin cos θ /sin θ /coscos θ /sin θ)] [∵ tan θ = sin θ /cos θ, cot θ = cos θ /sin θ]
    = [(sin θ /cos θ)/((sin θ – cos θ)/sin θ)]+[(cos θ /sin θ)/((cos θ – sin θ)/cos θ)]
    = sin² θ /(cos θ(sin θ – cos θ)) + cos² θ /(sin θ(cos θ – sin θ))
    = sin² θ /(cos θ(sin θ – cos θ)) + cos² θ /(sin θ(sin θ – cos θ)) [∵ (cos θ – sin θ) = -(sin θ – cos θ)]
    = (sin³ θ – cos³ θ)/(cos θ sin θ (sin θ – cos θ))
    = ((sin θ – cos θ)(sin² θ + cos² θ + cos θ sin θ))/(cos θ sin θ (sin θ – cos θ)) [∵ a³ – b³ = (a – b)(a² + b² + ab)]
    = (1 + cos θ sin θ)(cos θ sin θ)
    = (1/cos θ sin θ) + (cos θ sin θ /cos θ sin θ)
    = sec θ cosec θ + 1 = RHS

    (iv) (1 + sec A)/(sec A) = (sin² A)/1 – cos A)
    LHS = (1 + sec A)(sec A)
    = (1 + 1/cos A)/(1/cos A) [∵ sec A = 1/cos A]
    = ((cos A + 1)/cos A)/(1/cos A)
    = (1 + cos A)/1
    = (1 + cos A)/1 × (1 – cos A)/(1 – cos A)
    = (1 – cos² A)/(1 – cos A)
    = (sin² A)/(1 – cos A) [∵ 1 – cos² A = sin² A]
    = RHS

    (v) (cos A – sin A + 1)/(cos A + sin A -1) = cosec A + cot A
    LHS = (cos A – sin A + 1)/(cos A + sin A – 1)
    = (cot A – 1 + cosec A)/(cot A + 1 – cosec A) [Dividing Numerator and Denominator by sin A]
    = (cot A + cosec A – (1))/(cot A + 1 – cosec A)]
    = (cot A + cosec A – (cosec A + cot A)(cosec A – cot A))/(cot A + 1 – cosec A) [∵ cosec² A – cot² A = 1]
    = (cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)/(cot A + 1 – cosec A)
    = (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
    = cot A + cosec A = RHS

    (vi) √((1 + sin A)/(1 – sin A)) = sec A + tan A
    LHS = √((1 + sin A)/(1 – sin A))
    = √((1 + sin A)/(1 – sin A) × (1 + sin A)/(1 + sin A)) = √((1 + sin A)²/(1 – sin A))
    = √((1 + sin A)²/(cos² A)) [∵ 1 – sin² A = cos² A]
    = (1 + sin A)/(cos A)
    = (1/cos A) + (sin A/cos A)
    = sec A + tan A = RHS

    (vii) (sin θ – 2 sin³ θ)/(2 cos³ θ – cos θ) = tan θ
    LHS = (sin θ – 2 sin³ θ)/(2 cos³ θ – cos θ)
    = (sin θ(1 – 2 sin² θ))/(cos θ (2 cos² θ – 1))
    = (sin θ(1 – 2 sin² θ))/(cos θ [2(1 – sin² θ) – 1]) [∵ cos²θ = 1 – sin²θ]
    = (sin θ(1 – 2 sin² θ))/(cos θ (2 – 2 cos² θ – 1))
    = (sin θ(1 – 2 sin² θ))/(cos θ (1 – 2 sin² θ))
    = sin θ /cos θ = tan θ = RHS

    (viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
    LHS = (sin A + cosec A)² + (cos A + sec A)²
    = sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A
    = (sin² A + cos² A) + cosec² A + 2 + sec² A + 2 [∵ cos A sec A = 1, sin A cosec A = 1]
    = 1 + (1 + cot² A) + 2 + (1 + tan² A) + 2 [∵ cosec² A = 1 + cot² A, sec² A = 1 + tan² A]
    = 7 + tan² A + cot² A = RHS

    (ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)
    LHS = (cosec A – sin A)(sec A – cos A)
    = (1/sin A – sin A)(1/cos A – cos A)
    = ((1 – sin² A)/sin A)((1 – cos² A)/cos A)
    = (cos² A / sin A)(sin ² A / cos A)
    = sin A cos A …(i)
    = RHS = 1/(tan A + cot A)
    = (1/((sin A / sin A) + (cos A / sin A)) = (1/((sin² A + cos² A)/(cos A sin A))) = (1/(1/cos A sin A))
    = cos A sin A …(ii)
    From equation (i) and (ii), we get
    LHS = RHS

    (x) (1 + tan² A)/(1 + cot² A)
    = sec² A / cosec² A [∵ cosec² A = 1 + cot² A, sec² A = 1 + tan² A]
    = (1/cos² A)/(1/sin² A) = 1/(cos² A) × (sin² A)/1 = tan² A = RHS
    Now, ((1 – tan A)/(1 – cot A))²
    = ((1 – ((sin A)/(cos A)))/(1 – ((cos A)/(sin A))) = (((cos A – sin A)/(cos A))/((sin A – cos A)/(sin A)))²
    = ((cos A – sin A)/(cos A) × (sin A)/(sin A – cos A))² = (- (sin A – cos A)/(cos A) × (sin A)/(sin A – cos A))²
    = (- sin A / cos A)² = tan² A = RHS

    Here is the video solution 😄👇

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