Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p
Question 1 Report
Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p
Answer Details
We know that \(\sin p = \dfrac{5}{13}\) and $p$ is acute.
Using Pythagorean identity, we can find $\cos p$:
\begin{align*}
\cos^2 p &= 1 - \sin^2 p \\
\cos^2 p &= 1 - \left(\dfrac{5}{13}\right)^2 \\
\cos^2 p &= \dfrac{144}{169} \\
\cos p &= \dfrac{12}{13}
\end{align*}
Using the definition of tangent, we can find $\tan p$:
\begin{align*}
\tan p &= \dfrac{\sin p}{\cos p} \\
\tan p &= \dfrac{\frac{5}{13}}{\frac{12}{13}} \\
\tan p &= \dfrac{5}{12}
\end{align*}
Therefore,
\begin{align*}
\cos p - \tan p &= \dfrac{12}{13} - \dfrac{5}{12} \\
&= \dfrac{144}{156} - \dfrac{65}{156} \\
&= \dfrac{79}{156}
\end{align*}
Hence, the answer is \(\frac{79}{156}\).
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