Show that the equation
\(\frac{\tan x + \sin x}{\tan x - \sin x} = k,\)
where \(k\) is a constant, may be expressed as
\(\frac{1 + \cos x}{1 - \cos x} = k.\)
Solution
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:
\(\frac{1 + \cos x}{1 - \cos x} = k.\)
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