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.

(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 😄👇

Given that: cot 0 = 7/8
Let cot 0=where k is a real number.
In △ABC, by Pythagoras theorem, we have
AC² = BC²+AB²
= (8k)²+(7k)²
= 64k²+49k²
= 113k²
⇒ AC = √(113k²) = √(113k)
(i) ((1+sin θ)(1-sin 9))/((1+cos θ)(1-cos θ))
= ((1+ 8/√113)(1- 8/√113))/((1+ 7√113)(1- 7/√113))
= (1-(8/√113)²)/(1-(7/√113)²) = (1- 64/113)/(1- 49/113) = ((113-64)/113)/((113-49)/113) = 49/64
(ii) cot² θ = (cot θ)² = (7/8)² = 49/64

Here is the Video explanation of the above question😀👇

Given that: cos A= cos B
cos A = cos B
⇒ AP/AQ = BC/BD
⇒ AP/BC = AQ/BD
Let AP/BC = AQ/BC = k …(i)
Therefore, AP = k(BC) and AQ = k(BD)
Now, In △APQ and △BCD
PQ/CD = √((AQ²-AP²))/(√(BD²-BC²))
= (√((k.BD)²-(k.BC)²))/(√(BD²-BC²))
= (k√(BD²-BC²))/(√(BD²-BC²)) = k …(ii)
From the equation (i) and (ii), we get
AP/BC = AQ/BD = PQ/CD
So, △APQ ∼ △BCD [SSS similarity criteria]
Hence, ∠A = ∠B

Video explanation of this question👌🤗