Question

\[\frac{1 - \sin A}{\cos A} = \sec A - \tan A.\]

Statement (1): \[\frac{1 - \sin A}{\cos A} = \sec A - \tan A\]

\[\Rightarrow \frac{1 + \sin A}{\cos A} = \sec A + \tan A\]

Statement (2): \[\frac{1 - \sin A}{\cos A} - \frac{1 + \sin A}{\cos A} = 2\sec A\]

Options

• Both the statements are true.

• Both the statements are false.

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

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

Answer 1

Statement 1 is true, and statement 2 is false.

Explanation:

Given, \[\frac{1 - \sin A}{\cos A} = \sec A - \tan A\]

Taking L.H.S.

\[\Rightarrow \frac{1 - \sin A}{\cos A}\]

\[\Rightarrow \frac{1}{\cos A} - \frac{\sin A}{\cos A}\]

\[\Rightarrow \sec A - \tan A\]

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

Now, \[\frac{1 + \sin A}{\cos A} = \sec A + \tan A\]

Taking L.H.S.

\[\Rightarrow \frac{1 + \sin A}{\cos A}\]

\[\Rightarrow \frac{1}{\cos A} + \frac{\sin A}{\cos A}\]

\[\Rightarrow \sec A + \tan A\]

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

So, statement 1 is true.

Now, \[\frac{1 - \sin A}{\cos A} - \frac{1 + \sin A}{\cos A} = 2\sec A\]

Solving L.H.S.

\[\Rightarrow \frac{1 - \sin A}{\cos A} - \frac{1 + \sin A}{\cos A}\]

\[\Rightarrow \frac{1}{\cos A} - \frac{\sin A}{\cos A} - \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)\]

⇒ (sec A − tan A) − (sec A + tan A)

⇒ sec A − tan A − sec A v 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.