(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👌🤗