$\dfrac{1 - \text{sin A}}{\text{cos A}}$ = sec A - tan A

Statement (1): $\dfrac{1 - \text{sin A}}{\text{cos A}}$ = sec A - tan A

$\Rightarrow \dfrac{1 + \text{sin A}}{\text{cos A}}$ = sec A + tan A

Statement (2): $\dfrac{1 - \text{sin A}}{\text{cos A}} - \dfrac{1 + \text{sin A}}{\text{cos A}}$ = 2sec A

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Given, $\dfrac{1 - \text{sin A}}{\text{cos A}}$ = sec A - tan A

Taking L.H.S.

$\Rightarrow \dfrac{1 - \text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \text{sec A} - \text{tan A}$

As, L.H.S. = R.H.S., so equation is true.

Now, $\dfrac{1 + \text{sin A}}{\text{cos A}}$ = sec A + tan A

Taking L.H.S.

$\Rightarrow \dfrac{1 + \text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \text{sec A} + \text{tan A}$

As, L.H.S. = R.H.S., so equation is true.

So, statement 1 is true.

Now, $\dfrac{1 - \text{sin A}}{\text{cos A}} - \dfrac{1 + \text{sin A}}{\text{cos A}}$ = 2sec A

Solving L.H.S.

$\Rightarrow \dfrac{1 - \text{sin A}}{\text{cos A}} - \dfrac{1 + \text{sin A}}{\text{cos A}}$

$\Rightarrow \dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}} - \Big(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big)$

⇒ (sec A - tan A) - (sec A + tan A)

⇒ sec A - tan A - sec A - tan A

⇒ -2tan A.

As, L.H.S. ≠ R.H.S.

So, statement 2 is false.

∴ Statement 1 is true and statement 2 is false.

Hence, option 3 is the correct option.