What is the value of `[ tan (90 - A) + cot (90 - A) ] ^(2) // [ 2 sec^(2) (90 - 2A)]` ?

Options

A. 0
B. 1
C. 2
D. `-1`

Answer

To solve the given expression \(\frac{[ \tan(90^\circ - A) + \cot(90^\circ - A) ]^2}{2 \sec^2(90^\circ - 2A)}\), we will simplify it step by step.

### Step 1: Simplify \(\tan(90^\circ - A)\) and \(\cot(90^\circ - A)\)

Using the co-function identities:
\[ \tan(90^\circ - A) = \cot(A) \]
\[ \cot(90^\circ - A) = \tan(A) \]

Thus, we can rewrite the expression:
\[ \tan(90^\circ - A) + \cot(90^\circ - A) = \cot(A) + \tan(A) \]

### Step 2: Rewrite the expression

Now we can substitute this back into the original expression:
\[ \frac{[ \cot(A) + \tan(A) ]^2}{2 \sec^2(90^\circ - 2A)} \]

### Step 3: Simplify \(\sec^2(90^\circ - 2A)\)

Using the co-function identity:
\[ \sec(90^\circ - x) = \csc(x) \]
So,
\[ \sec^2(90^\circ - 2A) = \csc^2(2A) \]

### Step 4: Substitute back into the expression

Now we can substitute this into our expression:
\[ \frac{[ \cot(A) + \tan(A) ]^2}{2 \csc^2(2A)} \]

### Step 5: Simplify \(\cot(A) + \tan(A)\)

We know:
\[ \cot(A) = \frac{\cos(A)}{\sin(A)} \quad \text{and} \quad \tan(A) = \frac{\sin(A)}{\cos(A)} \]
Thus,
\[ \cot(A) + \tan(A) = \frac{\cos(A)}{\sin(A)} + \frac{\sin(A)}{\cos(A)} = \frac{\cos^2(A) + \sin^2(A)}{\sin(A) \cos(A)} = \frac{1}{\sin(A) \cos(A)} \]

### Step 6: Substitute back into the expression

Now we can substitute this back into our expression:
\[ \frac{[ \frac{1}{\sin(A) \cos(A)} ]^2}{2 \csc^2(2A)} \]

### Step 7: Simplify further

Calculating the square:
\[ \frac{1}{\sin^2(A) \cos^2(A)} \text{ and } \csc^2(2A) = \frac{1}{\sin^2(2A)} \]
We know \(\sin(2A) = 2 \sin(A) \cos(A)\), so:
\[ \csc^2(2A) = \frac{1}{4 \sin^2(A) \cos^2(A)} \]

### Step 8: Final substitution

Now substituting into the expression:
\[ \frac{\frac{1}{\sin^2(A) \cos^2(A)}}{2 \cdot \frac{1}{4 \sin^2(A) \cos^2(A)}} = \frac{1}{\sin^2(A) \cos^2(A)} \cdot \frac{4 \sin^2(A) \cos^2(A)}{2} = \frac{4}{2} = 2 \]

### Final Answer

Thus, the value of the expression is:
\[ \boxed{2} \]