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In a right triangle $ABC$, right-angled at $B$, if $ an A=\frac{4}{3}$, then $ rac{\sin A}{\cos A}$ is equal to
A$\frac{3}{4}$
B$\frac{4}{3}$
C$\frac{5}{3}$
D$1$
Answer & Solution
Correct answer: B. $\frac{4}{3}$
Using the identity $\tan A=\frac{\sin A}{\cos A}$, we get $\frac{\sin A}{\cos A}=\tan A=\frac{4}{3}$. So the correct option is $\frac{4}{3}$.
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