If \(\sin x = \dfrac{3}{5}\) and \(x \in \left(0, \dfrac{\pi}{2}\right)\), then find \(\dfrac{1 + \tan x}{1 - \tan x}\).
7
Step 1 – find \(\cos x\) and \(\tan x\):
Since \(x\) lies in the first quadrant, all ratios are positive.
\(\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \dfrac{9}{25}} = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}\)
\(\tan x = \dfrac{\sin x}{\cos x} = \dfrac{3/5}{4/5} = \dfrac{3}{4}\)
Step 2 – substitute into the expression:
\(\dfrac{1 + \tan x}{1 - \tan x} = \dfrac{1 + \tfrac{3}{4}}{1 - \tfrac{3}{4}} = \dfrac{7/4}{1/4} = 7\)
Hence the answer is 7.
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