If `cosec x + cot x = a`, then `cos x` is

Options

A. `(a-1)/( a^2 + 1)`
B. `(a^2 - 1)/( a^2 + 1)`
C. `(2a)/( a^2 + 1)`
D. `(2a)/( a^2 -1)`

Answer

To find the value of \( \cos x \) given that \( \csc x + \cot x = a \), we can follow these steps:

### Step 1: Express \( \csc x \) and \( \cot x \) in terms of \( \sin x \) and \( \cos x \)
We know that:
\[ \csc x = \frac{1}{\sin x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} \]
Thus, we can rewrite the equation:
\[ \csc x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1 + \cos x}{\sin x} \]
Setting this equal to \( a \):
\[ \frac{1 + \cos x}{\sin x} = a \]

### Step 2: Rearranging the equation
From the equation above, we can express \( 1 + \cos x \):
\[ 1 + \cos x = a \sin x \]

### Step 3: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \sin^2 x + \cos^2 x = 1 \]
We can express \( \sin x \) in terms of \( \cos x \):
\[ \sin^2 x = 1 - \cos^2 x \]

### Step 4: Substitute \( \sin x \) in the equation
Substituting \( \sin x = \sqrt{1 - \cos^2 x} \) into the equation \( 1 + \cos x = a \sin x \):
\[ 1 + \cos x = a \sqrt{1 - \cos^2 x} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (1 + \cos x)^2 = a^2 (1 - \cos^2 x) \]
Expanding both sides:
\[ 1 + 2\cos x + \cos^2 x = a^2 - a^2 \cos^2 x \]

### Step 6: Rearranging the equation
Combine like terms:
\[ \cos^2 x + a^2 \cos^2 x + 2\cos x + 1 - a^2 = 0 \]
Factoring out \( \cos^2 x \):
\[ (1 + a^2) \cos^2 x + 2\cos x + (1 - a^2) = 0 \]

### Step 7: Use the quadratic formula
This is a quadratic equation in terms of \( \cos x \):
\[ A = 1 + a^2, \quad B = 2, \quad C = 1 - a^2 \]
Using the quadratic formula:
\[ \cos x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Substituting the values:
\[ \cos x = \frac{-2 \pm \sqrt{2^2 - 4(1 + a^2)(1 - a^2)}}{2(1 + a^2)} \]
\[ = \frac{-2 \pm \sqrt{4 - 4(1 + a^2)(1 - a^2)}}{2(1 + a^2)} \]
\[ = \frac{-2 \pm \sqrt{4 - 4(1 - a^4)}}{2(1 + a^2)} \]
\[ = \frac{-2 \pm \sqrt{4a^4}}{2(1 + a^2)} \]
\[ = \frac{-2 \pm 2a^2}{2(1 + a^2)} \]
\[ = \frac{-1 \pm a^2}{1 + a^2} \]

### Final Result
Thus, the value of \( \cos x \) is:
\[ \cos x = \frac{a^2 - 1}{a^2 + 1} \]