If $\sin A =\frac{3}{4}$ calculate $\cos A$ and $\tan A$.

Let we have a triangle, 

Since $Sin \left(A\right) =\frac{Opposite side}{Hypotenuse}= \frac{BC}{AC}=\frac{3}{4}$

Let BC be 3k and AC will be 4k

where k is a positive real number.

Using Pythagoras theorem, we get

$A{C}^2=A{B}^2 + B{C}^2$

${\left(4k\right)}^2=A{B}^2 + {\left(3k\right)}^2$

$16k^2-9k^2 =A{B}^2$

$A{B}^2=7k^2$

Thus, $AB =\sqrt{7}k$

$Cos \left(A\right) =\frac{Adjacent side}{Hypotenuse}= \frac{AB}{AC}= \frac{\sqrt{7}k}{4k}=\frac{\sqrt{7}}{4}$

$\tan \left(A\right) =\frac{Opposite side}{Adjacent side}=\frac{BC}{AB}=\frac{3k}{\sqrt{7}k}=\frac{3}{\sqrt{7}}$