If `(cotA-cosecA)^(2)=x` , then the value of x is

Options

A. `((1-cosA))/((1+cosA))`
B. `((1-sinA))/((1+sinA))`
C. `((1-secA))/((1+secA))`
D. `((1-cosecA))/((1+cosecA))`

Answer

To solve the problem `(cotA - cosecA)² = x`, we will follow these steps:

### Step 1: Rewrite cotangent and cosecant in terms of sine and cosine
We know that:
- \( \cot A = \frac{\cos A}{\sin A} \)
- \( \csc A = \frac{1}{\sin A} \)

Substituting these into the expression gives us:
\[ x = \left( \frac{\cos A}{\sin A} - \frac{1}{\sin A} \right)^2 \]

### Step 2: Combine the terms inside the parentheses
Since both terms have a common denominator of \( \sin A \), we can combine them:
\[ x = \left( \frac{\cos A - 1}{\sin A} \right)^2 \]

### Step 3: Square the fraction
Now we can square the fraction:
\[ x = \frac{(\cos A - 1)^2}{\sin^2 A} \]

### Step 4: Use the Pythagorean identity
Recall the Pythagorean identity:
\[ \sin^2 A + \cos^2 A = 1 \implies \sin^2 A = 1 - \cos^2 A \]
Thus, we can rewrite \( \sin^2 A \) in the denominator:
\[ x = \frac{(\cos A - 1)^2}{1 - \cos^2 A} \]

### Step 5: Factor the denominator
The denominator \( 1 - \cos^2 A \) can be factored using the difference of squares:
\[ 1 - \cos^2 A = (1 - \cos A)(1 + \cos A) \]
So we can substitute this back into our expression for \( x \):
\[ x = \frac{(\cos A - 1)^2}{(1 - \cos A)(1 + \cos A)} \]

### Step 6: Simplify the expression
Notice that \( \cos A - 1 = -(1 - \cos A) \), so we can rewrite \( (\cos A - 1)^2 \) as:
\[ (\cos A - 1)^2 = (-(1 - \cos A))^2 = (1 - \cos A)^2 \]
Thus, we have:
\[ x = \frac{(1 - \cos A)^2}{(1 - \cos A)(1 + \cos A)} \]
Now, we can cancel one \( (1 - \cos A) \) from the numerator and the denominator:
\[ x = \frac{1 - \cos A}{1 + \cos A} \]

### Final Result
Thus, the value of \( x \) is:
\[ x = \frac{1 - \cos A}{1 + \cos A} \]

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