(i) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

     [Hint : Write the expression in terms of sin θ and cos θ]

(ii) (1 + sec A)/sec A = sin²A/(1-cos A)  

     [Hint : Simplify LHS and RHS separately]

Solutions:

(i) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)

We all know that tan θ =sin θ/cos θ

cot θ = cos θ/sin θ

Now, substitute it in the given equation to simplify it.

= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]

= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]

= sin²θ/[cos θ(sin θ-cos θ)] + cos²θ/[sin θ(cos θ-sin θ)]

= sin²θ/[cos θ(sin θ-cos θ)] – cos²θ/[sin θ(sin θ-cos θ)]

= 1/(sin θ-cos θ) [(sin²θ/cos θ) – (cos²θ/sin θ)]

= 1/(sin θ-cos θ) × [(sin³θ – cos³θ)/sin θ cos θ]

= [(sin θ-cos θ)(sin²θ+cos²θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]

= (1 + sin θ cos θ)/sin θ cos θ

= 1/sin θ cos θ + 1

= 1 + sec θ cosec θ = R.H.S.

So, L.H.S. = R.H.S.

As a result, hence proved

(ii)  (1 + sec A)/sec A = sin²A/(1-cos A)

To simplify the above problem, we first need to simplify the L.H.S. side

L.H.S. = (1 + sec A)/sec A

Because the secant function is the inverse of the cos function and is denoted by

= (1 + 1/cos A)/1/cos A

= (cos A + 1)/cos A/1/cos A

As a result, (1 + sec A)/sec A = cos A + 1

R.H.S. = sin²A/(1-cos A)

We all know that sin²A = (1 – cos²A), as a result we get,

= (1 – cos²A)/(1-cos A)

= (1-cos A)(1+cos A)/(1-cos A)

Therefore, sin²A/(1-cos A)= cos A + 1

L.H.S. = R.H.S.

As a result, hence proved.