If $cos\ A = \frac{4}{5}$, then the value of $tan\ A$ is
(A) $\frac{3}{5}$
(B) $\frac{3}{4}$
(C) $\frac{4}{3}$
(D) $\frac{5}{3}$

Given:

$cos\ A = \frac{4}{5}$

To do:

We have to find the value of $tan\ A$.

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $cos\ A=\frac{4}{5}$.

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$sec\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$

$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow (5)^2=(4)^2+BC^2$

$\Rightarrow BC^2=25-16$

$\Rightarrow BC=\sqrt{9}=3$

Therefore,

$tan\ A=\frac{BC}{AB}=\frac{3}{4}$.