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

Let us draw a right angle triangle ABC.

We know that,

sin A = $\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}}$

Substituting values we get,

$\dfrac{3}{4} = \dfrac{BC}{AC}$

Let BC = 3k and AC = 4k.

In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ (4k)² = AB² + (3k)²

⇒ AB² = 16k² - 9k²

⇒ AB² = 7k²

⇒ AB = $\sqrt{7k^2} = \sqrt{7}k$

We know that,

cos A = $\dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{\sqrt{7}k}{4k} = \dfrac{\sqrt{7}}{4}$.

tan A = $\dfrac{\text{Side opposite to ∠A}}{\text{Side adjacent to ∠A}} = \dfrac{BC}{AB} = \dfrac{3k}{\sqrt{7}k} = \dfrac{3}{\sqrt{7}}$.

Hence, $\text{cos A} = \dfrac{\sqrt{7}}{4} \text{ and tan A} = \dfrac{3}{\sqrt{7}}$.