If sin x = 5/13 and x lies in (0°, 90°), then find the value of: \(\dfrac{1 + \cos x}{1 - \cos x}\)
25
We are given \(\sin x = \dfrac{5}{13}\) with \(x\) in \((0\degree, 90\degree)\), so \(x\) is acute and \(\cos x\) is positive.
Using \(\sin^2 x + \cos^2 x = 1\), \(\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \dfrac{25}{169}} = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13}\).
Compute the numerator: \(1 + \cos x = 1 + \dfrac{12}{13} = \dfrac{25}{13}\).
Compute the denominator: \(1 - \cos x = 1 - \dfrac{12}{13} = \dfrac{1}{13}\).
Divide: \(\dfrac{1 + \cos x}{1 - \cos x} = \dfrac{25/13}{1/13} = \dfrac{25}{13}\times\dfrac{13}{1} = 25\).
Key concept: for an acute angle take the positive square root when recovering \(\cos x\), then simplify the fractions before dividing.
Hence the value is 25.
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