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

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

Simplify both sides.

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

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

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

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

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 $\frac{\cos (x)}{\cos (x)}$ .

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

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

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

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

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

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

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

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

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

Solve the equation for $B$ .

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

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

Divide each term in $(B + \cos (x)) (1 - \cos (x)) = (A - \cos (x)) (1 + \cos (x))$ by $1 - \cos (x)$ and simplify.

Divide each term in $(B + \cos (x)) (1 - \cos (x)) = (A - \cos (x)) (1 + \cos (x))$ by $1 - \cos (x)$ .

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

Simplify the left side.

Cancel the common factor of $1 - \cos (x)$ .

Cancel the common factor.

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

Divide $B + \cos (x)$ by $1$ .

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

Subtract $\cos (x)$ from both sides of the equation.

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

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

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