(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
Solution:
(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
Take the Left-Hand Side (L.H.S) of the given equation to prove the Right-Hand Side (R.H.S)
L.H.S. = (cosec θ – cot θ)2
Expand the given equation, which is written as (a-b)2.
As we know (a-b)2 = a2 + b2 – 2ab
Here a = cosec θ and b = cot θ
= (cosec2θ + cot2θ – 2cosec θ cot θ)
To simplify, we need to apply the equivalent ratios and the corresponding inverse functions
= (1/sin2θ + cos2θ/sin2θ – 2cos θ/sin2θ)
= (1 + cos2θ – 2cos θ)/(1 – cos2θ)
= (1-cos θ)2/(1 – cosθ)(1+cos θ)
= (1-cos θ)/(1+cos θ) = R.H.S.
As a result, (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
Hence proved.
(ii) (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A
We need to take the Left Hand Side(L.H.S) of the equation
L.H.S. = (cos A/(1+sin A)) + ((1+sin A)/cos A)
= [cos2A + (1+sin A)2]/(1+sin A)cos A
= (cos2A + sin2A + 1 + 2sin A)/(1+sin A) cos A
As we know, cos2A + sin2A = 1, so it can be written as,
= (1 + 1 + 2sin A)/(1+sin A) cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A = 2 sec A = Right Hand Side (R.H.S.)
L.H.S. = R.H.S.
(cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A
As a result, hence proved.