Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p
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Correct Answer: Option A
Explanation:
If \(sin P = \frac{5}{13}\) from right angled triangle from pythagoras theorem
\(BC^2 = 13^2 - 5^2\\
=169-25\\
BC = \sqrt{144} = 12\\
∴ cos P - tan P = \frac{12}{13} - \frac{5}{12}\\
=\frac{79}{156}\)
If \(sin P = \frac{5}{13}\) from right angled triangle from pythagoras theorem
\(BC^2 = 13^2 - 5^2\\
=169-25\\
BC = \sqrt{144} = 12\\
∴ cos P - tan P = \frac{12}{13} - \frac{5}{12}\\
=\frac{79}{156}\)