If tan x = \(\frac{4}{3}\), 0\(^o\) < x < 90\(^o\), find the value of sin x - cos x
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Correct Answer: Option B
Explanation:
From the diagram,
h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras')
h\(^2\) = 16 + 9 = 25
h = \(\sqrt{25}\) = 5
Hence, sin x - cos x
= \(\frac{4}{5} - \frac{3}{5}\)
= \(\frac{2}{5}\)
From the diagram,
h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras')
h\(^2\) = 16 + 9 = 25
h = \(\sqrt{25}\) = 5
Hence, sin x - cos x
= \(\frac{4}{5} - \frac{3}{5}\)
= \(\frac{2}{5}\)