If sin x = \(\frac{5}{13}\) and 0^o \(\leq\) x \(\leq\) 90^o, find the value of (cos x - tan x)

\(\frac{7}{13}\)

\(\frac{12}{13}\)

\(\frac{79}{156}\)

\(\frac{209}{156}\)

Correct answer is C

Sin x = \(\frac{5}{13}\)

0^o \(\leq\) x \(\leq\) 90^o, (cos x - tan x)

AC² = AB² + BC²

13² = 5² + BC²

169 - 25 + BC²

169 - 25 = BC²

144 = BC²

Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\)

BC = \(\sqrt{144}\)

BC = 12

tan x = \(\frac{opp}{adj} = \frac{5}{12}\)

BC = 12

cos x - tan x = \(\frac{12}{13} - \frac{5}{12}\)

\(\frac{144 - 65}{156} = \frac{79}{156}\)

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If sin x = \(\frac{5}{13}\) and 0o \(\leq\) x \(\leq\) 90o, find the value of (cos x - tan x)

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If sin x = \(\frac{5}{13}\) and 0^o \(\leq\) x \(\leq\) 90^o, find the value of (cos x - tan x)