Prove that
`(cos A )/( 1 - sin A ) = tan (45 ^(@) + (A)/(2)),`

Answer

To prove that

\[ \frac{\cos A}{1 - \sin A} = \tan\left(45^\circ + \frac{A}{2}\right), \]

we will start with the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step 1: Write down the LHS

We start with the left-hand side:

\[ \text{LHS} = \frac{\cos A}{1 - \sin A}. \]

### Step 2: Use the half-angle identities

Recall the half-angle identities:

\[ \cos A = \frac{1 - \tan^2\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)}, \]
\[ \sin A = \frac{2\tan\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)}. \]

Substituting these into our LHS gives:

\[ \text{LHS} = \frac{\frac{1 - \tan^2\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)}}{1 - \frac{2\tan\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)}}. \]

### Step 3: Simplify the denominator

To simplify the denominator, we can rewrite it as:

\[ 1 - \frac{2\tan\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)} = \frac{(1 + \tan^2\left(\frac{A}{2}\right)) - 2\tan\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)} = \frac{(1 - \tan\left(\frac{A}{2}\right))^2}{1 + \tan^2\left(\frac{A}{2}\right)}. \]

### Step 4: Substitute back into the LHS

Now substituting this back into the LHS gives:

\[ \text{LHS} = \frac{1 - \tan^2\left(\frac{A}{2}\right)}{(1 + \tan^2\left(\frac{A}{2}\right))} \cdot \frac{1 + \tan^2\left(\frac{A}{2}\right)}{(1 - \tan\left(\frac{A}{2}\right))^2}. \]

The \(1 + \tan^2\left(\frac{A}{2}\right)\) cancels out:

\[ \text{LHS} = \frac{1 - \tan^2\left(\frac{A}{2}\right)}{(1 - \tan\left(\frac{A}{2}\right))^2}. \]

### Step 5: Factor the numerator

The numerator can be factored:

\[ 1 - \tan^2\left(\frac{A}{2}\right) = (1 - \tan\left(\frac{A}{2}\right))(1 + \tan\left(\frac{A}{2}\right)). \]

Thus, we have:

\[ \text{LHS} = \frac{(1 - \tan\left(\frac{A}{2}\right))(1 + \tan\left(\frac{A}{2}\right))}{(1 - \tan\left(\frac{A}{2}\right))^2}. \]

### Step 6: Cancel common terms

Canceling \(1 - \tan\left(\frac{A}{2}\right)\) from the numerator and denominator gives:

\[ \text{LHS} = \frac{1 + \tan\left(\frac{A}{2}\right)}{1 - \tan\left(\frac{A}{2}\right)}. \]

### Step 7: Recognize the tangent addition formula

This expression can be recognized as the tangent addition formula:

\[ \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}. \]

Setting \(x = 45^\circ\) (where \(\tan 45^\circ = 1\)) and \(y = \frac{A}{2}\):

\[ \tan\left(45^\circ + \frac{A}{2}\right) = \frac{1 + \tan\left(\frac{A}{2}\right)}{1 - \tan\left(\frac{A}{2}\right)}. \]

### Conclusion

Thus, we have shown that:

\[ \frac{\cos A}{1 - \sin A} = \tan\left(45^\circ + \frac{A}{2}\right), \]

which completes the proof.

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