What is the value of ` [ 1 - sin ( 90 - 2A ) ] // [ 1 + sin ( 90 + 2 A ) ] `?

Options

A. sin A cos A
B. ` cot^(2) A `
C. ` tan ^(2) A `
D. `sin ^(2) A cos A `

Answer

To solve the expression \( \frac{1 - \sin(90^\circ - 2A)}{1 + \sin(90^\circ + 2A)} \), we can follow these steps:

### Step 1: Simplify the sine functions
Using the trigonometric identities:
- \( \sin(90^\circ - \theta) = \cos(\theta) \)
- \( \sin(90^\circ + \theta) = \cos(\theta) \)

We can rewrite the expression:
\[ \sin(90^\circ - 2A) = \cos(2A) \]
\[ \sin(90^\circ + 2A) = \cos(2A) \]

Thus, the expression becomes:
\[ \frac{1 - \cos(2A)}{1 + \cos(2A)} \]

### Step 2: Use the identity for \( 1 - \cos(2A) \) and \( 1 + \cos(2A) \)
We know the following identities:
- \( 1 - \cos(2A) = 2 \sin^2(A) \)
- \( 1 + \cos(2A) = 2 \cos^2(A) \)

Substituting these identities into the expression gives:
\[ \frac{2 \sin^2(A)}{2 \cos^2(A)} \]

### Step 3: Simplify the fraction
The \( 2 \) in the numerator and denominator cancels out:
\[ \frac{\sin^2(A)}{\cos^2(A)} = \tan^2(A) \]

### Final Result
Thus, the value of the expression \( \frac{1 - \sin(90^\circ - 2A)}{1 + \sin(90^\circ + 2A)} \) is:
\[ \boxed{\tan^2(A)} \]