Solve for A (sec(x)-1)/(sec(x)+1)=(A-cos(x))/(B+cos(x))

$\frac{\sec (x) - 1}{\sec (x) + 1} = \frac{A - \cos (x)}{B + \cos (x)}$

Rewrite the equation as $\frac{A - \cos (x)}{B + \cos (x)} = \frac{\sec (x) - 1}{\sec (x) + 1}$ .

$\frac{A - \cos (x)}{B + \cos (x)} = \frac{\sec (x) - 1}{\sec (x) + 1}$

Multiply both sides by $B + \cos (x)$ .

$\frac{A - \cos (x)}{B + \cos (x)} (B + \cos (x)) = \frac{\sec (x) - 1}{\sec (x) + 1} (B + \cos (x))$

Simplify.

Simplify the left side.

Cancel the common factor of $B + \cos (x)$ .

Cancel the common factor.

$\frac{A - \cos (x)}{B + \cos (x)} (B + \cos (x)) = \frac{\sec (x) - 1}{\sec (x) + 1} (B + \cos (x))$

Rewrite the expression.

$A - \cos (x) = \frac{\sec (x) - 1}{\sec (x) + 1} (B + \cos (x))$

Simplify the right side.

Simplify $\frac{\sec (x) - 1}{\sec (x) + 1} (B + \cos (x))$ .

Rewrite $\sec (x)$ in terms of sines and cosines.

$A - \cos (x) = \frac{\frac{1}{\cos (x)} - 1}{\sec (x) + 1} (B + \cos (x))$

Rewrite $\sec (x)$ in terms of sines and cosines.

$A - \cos (x) = \frac{\frac{1}{\cos (x)} - 1}{\frac{1}{\cos (x)} + 1} (B + \cos (x))$

Multiply the numerator and denominator of the complex fraction by $\cos (x)$ .

Multiply $\frac{\frac{1}{\cos (x)} - 1}{\frac{1}{\cos (x)} + 1}$ by $\frac{\cos (x)}{\cos (x)}$ .

$A - \cos (x) = \frac{\cos (x)}{\cos (x)} \cdot \frac{\frac{1}{\cos (x)} - 1}{\frac{1}{\cos (x)} + 1} (B + \cos (x))$

Combine.

$A - \cos (x) = \frac{\cos (x) (\frac{1}{\cos (x)} - 1)}{\cos (x) (\frac{1}{\cos (x)} + 1)} (B + \cos (x))$

Apply the distributive property.

$A - \cos (x) = \frac{\cos (x) \frac{1}{\cos (x)} + \cos (x) \cdot - 1}{\cos (x) \frac{1}{\cos (x)} + \cos (x) \cdot 1} (B + \cos (x))$

Simplify by cancelling.

Cancel the common factor of $\cos (x)$ .

Cancel the common factor.

$A - \cos (x) = \frac{\cos (x) \frac{1}{\cos (x)} + \cos (x) \cdot - 1}{\cos (x) \frac{1}{\cos (x)} + \cos (x) \cdot 1} (B + \cos (x))$

Rewrite the expression.

$A - \cos (x) = \frac{1 + \cos (x) \cdot - 1}{\cos (x) \frac{1}{\cos (x)} + \cos (x) \cdot 1} (B + \cos (x))$

Cancel the common factor of $\cos (x)$ .

Cancel the common factor.

$A - \cos (x) = \frac{1 + \cos (x) \cdot - 1}{\cos (x) \frac{1}{\cos (x)} + \cos (x) \cdot 1} (B + \cos (x))$

Rewrite the expression.

$A - \cos (x) = \frac{1 + \cos (x) \cdot - 1}{1 + \cos (x) \cdot 1} (B + \cos (x))$

Simplify the numerator.

Move $- 1$ to the left of $\cos (x)$ .

$A - \cos (x) = \frac{1 - 1 \cdot \cos (x)}{1 + \cos (x) \cdot 1} (B + \cos (x))$

Rewrite $- 1 \cos (x)$ as $- \cos (x)$ .

$A - \cos (x) = \frac{1 - \cos (x)}{1 + \cos (x) \cdot 1} (B + \cos (x))$

Simplify terms.

Multiply $\cos (x)$ by $1$ .

$A - \cos (x) = \frac{1 - \cos (x)}{1 + \cos (x)} (B + \cos (x))$

Apply the distributive property.

$A - \cos (x) = \frac{1 - \cos (x)}{1 + \cos (x)} B + \frac{1 - \cos (x)}{1 + \cos (x)} \cos (x)$

Combine $\frac{1 - \cos (x)}{1 + \cos (x)}$ and $B$ .

$A - \cos (x) = \frac{(1 - \cos (x)) B}{1 + \cos (x)} + \frac{1 - \cos (x)}{1 + \cos (x)} \cos (x)$

Combine $\frac{1 - \cos (x)}{1 + \cos (x)}$ and $\cos (x)$ .

$A - \cos (x) = \frac{(1 - \cos (x)) B}{1 + \cos (x)} + \frac{(1 - \cos (x)) \cos (x)}{1 + \cos (x)}$

Combine the numerators over the common denominator.

$A - \cos (x) = \frac{(1 - \cos (x)) B + (1 - \cos (x)) \cos (x)}{1 + \cos (x)}$

Factor $1 - \cos (x)$ out of $(1 - \cos (x)) B + (1 - \cos (x)) \cos (x)$ .

Factor $1 - \cos (x)$ out of $(1 - \cos (x)) B$ .

$A - \cos (x) = \frac{(1 - \cos (x)) (B) + (1 - \cos (x)) \cos (x)}{1 + \cos (x)}$

Factor $1 - \cos (x)$ out of $(1 - \cos (x)) \cos (x)$ .

$A - \cos (x) = \frac{(1 - \cos (x)) (B) + (1 - \cos (x)) (\cos (x))}{1 + \cos (x)}$

Factor $1 - \cos (x)$ out of $(1 - \cos (x)) (B) + (1 - \cos (x)) (\cos (x))$ .

$A - \cos (x) = \frac{(1 - \cos (x)) (B + \cos (x))}{1 + \cos (x)}$

Add $\cos (x)$ to both sides of the equation.

$A = \frac{(1 - \cos (x)) (B + \cos (x))}{1 + \cos (x)} + \cos (x)$

Solve for A (sec(x)-1)/(sec(x)+1)=(A-cos(x))/(B+cos(x))

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