Start with the given equation:

\(\frac{\tan x + \cos x}{\tan x - \cos x} = k\)

Multiply both sides by \(\tan x - \cos x\):

\(\tan x + \cos x = k(\tan x - \cos x)\)

Substitute \(\tan x = \frac{\sin x}{\cos x}\):

\(\frac{\sin x}{\cos x} + \cos x = k \left( \frac{\sin x}{\cos x} - \cos x \right)\)

Clear the fraction by multiplying through by \(\cos x\):

\(\sin x + \cos^2 x = k(\sin x - \cos^2 x)\)

Use \(\cos^2 x = 1 - \sin^2 x\):

\(\sin x + 1 - \sin^2 x = k(\sin x - (1 - \sin^2 x))\)

Simplify the equation:

\(\sin x + 1 - \sin^2 x = k \sin x - k + k \sin^2 x\)

Rearrange terms to gather like terms on one side:

\(k \sin^2 x + \sin^2 x + k \sin x - \sin x - k - 1 = 0\)

Combine like terms:

\((k+1) \sin^2 x + (k-1) \sin x - (k+1) = 0\)

This matches the required expression.

Start with the given equation:

\(\frac{\tan x + \sin x}{\tan x - \sin x} = k.\)

Express \(\tan x\) as \(\frac{\sin x}{\cos x}\):

\(\frac{\frac{\sin x}{\cos x} + \sin x}{\frac{\sin x}{\cos x} - \sin x} = k.\)

Multiply numerator and denominator by \(\cos x\):

\(\frac{\sin x + \sin x \cos x}{\sin x - \sin x \cos x} = k.\)

Factor out \(\sin x\) from both numerator and denominator:

\(\frac{\sin x (1 + \cos x)}{\sin x (1 - \cos x)} = k.\)

Cancel \(\sin x\) from numerator and denominator:

\(\frac{1 + \cos x}{1 - \cos x} = k.\)

Start with the left-hand side:

\(\frac{1 + \\sin x}{1 - \\sin x} - \frac{1 - \\sin x}{1 + \\sin x}\)

Combine over a common denominator:

\(\frac{(1 + \\sin x)^2 - (1 - \\sin x)^2}{(1 - \\sin x)(1 + \\sin x)}\)

Expand the numerators:

\(\frac{1 + 2 \\sin x + \\sin^2 x - (1 - 2 \\sin x + \\sin^2 x)}{1 - \\sin^2 x}\)

Simplify the numerator:

\(\frac{4 \\sin x}{1 - \\sin^2 x}\)

Since \(1 - \\sin^2 x = \\cos^2 x\), the expression becomes:

\(\frac{4 \\sin x}{\\cos^2 x}\)

Rewrite \(\\sin x / \\cos x\) as \(\\tan x\):

\(\frac{4 \\tan x}{\\cos x}\)

This matches the right-hand side, proving the identity.