Prove that : ` (cos ^(2) A + tan ^(2) A - 1 )/( sin ^(2) A ) = tan ^(2) A . `

Answer

To prove that

\[ \frac{\cos^2 A + \tan^2 A - 1}{\sin^2 A} = \tan^2 A, \]

we will follow these steps:

### Step 1: Rewrite \(\tan^2 A\)
Recall that \(\tan A = \frac{\sin A}{\cos A}\). Therefore,

\[ \tan^2 A = \frac{\sin^2 A}{\cos^2 A}. \]

### Step 2: Substitute \(\tan^2 A\) in the equation
Now, substitute \(\tan^2 A\) in the original equation:

\[ \frac{\cos^2 A + \frac{\sin^2 A}{\cos^2 A} - 1}{\sin^2 A}. \]

### Step 3: Simplify the numerator
The numerator becomes:

\[ \cos^2 A + \frac{\sin^2 A}{\cos^2 A} - 1. \]

To combine the terms, we can express \(\cos^2 A - 1\) as \(-\sin^2 A\) (using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\)). Thus, we have:

\[ \cos^2 A - 1 = -\sin^2 A. \]

So, the numerator can be rewritten as:

\[ -\sin^2 A + \frac{\sin^2 A}{\cos^2 A} = \frac{-\sin^2 A \cos^2 A + \sin^2 A}{\cos^2 A}. \]

### Step 4: Factor out \(\sin^2 A\)
Factoring out \(\sin^2 A\) from the numerator gives:

\[ \frac{\sin^2 A (1 - \cos^2 A)}{\cos^2 A}. \]

### Step 5: Use the identity \(1 - \cos^2 A = \sin^2 A\)
Now, we can replace \(1 - \cos^2 A\) with \(\sin^2 A\):

\[ \frac{\sin^2 A \cdot \sin^2 A}{\cos^2 A} = \frac{\sin^4 A}{\cos^2 A}. \]

### Step 6: Divide by \(\sin^2 A\)
Now, we return to our original equation:

\[ \frac{\frac{\sin^4 A}{\cos^2 A}}{\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A. \]

### Conclusion
Thus, we have shown that:

\[ \frac{\cos^2 A + \tan^2 A - 1}{\sin^2 A} = \tan^2 A. \]