The problem states that $\sin A = \frac{3}{4}$, and asks us to find the other trigonometric ratios for angle A.

1. Problem Description

The problem states that $\sin A = \frac{3}{4}$, and asks us to find the other trigonometric ratios for angle A.

2. Solution Steps

Since $\sin A = \frac{3}{4}$, we can think of this as the opposite side being 3 and the hypotenuse being 4 in a right-angled triangle.

We can use the Pythagorean theorem to find the adjacent side. Let the adjacent side be $x$.

Then, $x^2 + 3^2 = 4^2$.

$x^2 + 9 = 16$

$x^2 = 16 - 9$

$x^2 = 7$

$x = \sqrt{7}$

Now we can find the other trigonometric ratios:

$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{7}}{4}$

$\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{\sqrt{7}}$

$\csc A = \frac{1}{\sin A} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

$\sec A = \frac{1}{\cos A} = \frac{1}{\frac{\sqrt{7}}{4}} = \frac{4}{\sqrt{7}}$

$\cot A = \frac{1}{\tan A} = \frac{1}{\frac{3}{\sqrt{7}}} = \frac{\sqrt{7}}{3}$

3. Final Answer

$\cos A = \frac{\sqrt{7}}{4}$

$\tan A = \frac{3}{\sqrt{7}}$

$\csc A = \frac{4}{3}$

$\sec A = \frac{4}{\sqrt{7}}$

$\cot A = \frac{\sqrt{7}}{3}$