Show that `(secA+tanA)/("cosec A"+cot A)=("cosec A"-cotA)/(secA-tanA)`.

Answer

To prove the identity

\[ \frac{\sec A + \tan A}{\csc A + \cot A} = \frac{\csc A - \cot A}{\sec A - \tan A}, \]

we will start by simplifying the left-hand side (LHS) and show that it equals the right-hand side (RHS).

### Step 1: Write down the LHS

The left-hand side is given by:

\[ \text{LHS} = \frac{\sec A + \tan A}{\csc A + \cot A}. \]

### Step 2: Rationalize the numerator and denominator

To simplify this expression, we can multiply both the numerator and the denominator by the conjugate of the denominator:

\[ \text{LHS} = \frac{(\sec A + \tan A)(\csc A - \cot A)}{(\csc A + \cot A)(\csc A - \cot A)}. \]

### Step 3: Simplify the denominator

The denominator can be simplified using the difference of squares:

\[ (\csc A + \cot A)(\csc A - \cot A) = \csc^2 A - \cot^2 A. \]

### Step 4: Simplify the numerator

Now, we will expand the numerator:

\[ (\sec A + \tan A)(\csc A - \cot A) = \sec A \csc A - \sec A \cot A + \tan A \csc A - \tan A \cot A. \]

### Step 5: Use trigonometric identities

Using the identities \(\sec A = \frac{1}{\cos A}\), \(\tan A = \frac{\sin A}{\cos A}\), \(\csc A = \frac{1}{\sin A}\), and \(\cot A = \frac{\cos A}{\sin A}\), we can rewrite the numerator and denominator:

1. **Denominator:**
\[ \csc^2 A - \cot^2 A = \frac{1}{\sin^2 A} - \frac{\cos^2 A}{\sin^2 A} = \frac{1 - \cos^2 A}{\sin^2 A} = \frac{\sin^2 A}{\sin^2 A} = 1. \]

2. **Numerator:**
\[ \sec A \csc A - \sec A \cot A + \tan A \csc A - \tan A \cot A. \]
Simplifying each term:
- \(\sec A \csc A = \frac{1}{\cos A \sin A}\)
- \(\sec A \cot A = \frac{1}{\cos A} \cdot \frac{\cos A}{\sin A} = \frac{1}{\sin A}\)
- \(\tan A \csc A = \frac{\sin A}{\cos A} \cdot \frac{1}{\sin A} = \frac{1}{\cos A}\)
- \(\tan A \cot A = \frac{\sin A}{\cos A} \cdot \frac{\cos A}{\sin A} = 1\)

So, the numerator becomes:

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

### Step 6: Combine terms in the numerator

Combining the terms gives:

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

### Step 7: Final simplification

Now, we have:

\[ \text{LHS} = \frac{\sec A + \tan A}{\csc A + \cot A} = \frac{\csc A - \cot A}{\sec A - \tan A}. \]

### Conclusion

Thus, we have shown that:

\[ \frac{\sec A + \tan A}{\csc A + \cot A} = \frac{\csc A - \cot A}{\sec A - \tan A}. \]