WAEC SSCE - General Mathematics - 2001 (Objective)

Question 1 Report

Given that sin $P = \frac{5}{13}$, where p is acute, find the value of cos p - tan p

$\frac{79}{156}$ Correct answer

$\frac{85}{156}$

$\frac{7}{13}$

$\frac{8}{1}$

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 $\frac{79}{156}$.

Read lesson note on Sine, Cosine And Tangent Of An Angle (WAEC)






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