Let us assume a right angled triangle ABC, right angled at B

Given: Sin A = 3/4

Therefore, Sin A = Opposite side /Hypotenuse= 3/4

Let BC be 3k and AC will be 4k

where k is a positive real number.

AC² = AB² + BC²

Substitute the value of AC and BC

(4k)²= AB² + (3k)²

16k²− 9k² = AB²

AB²=7k²

AB = √7k

Now, we have to find the value of cos A and tan A

We know that,

Cos (A) = Adjacent side/Hypotenuse

Substitute the value of AB and AC we get,

AB/AC = √7k/4k = √7/4

cos (A) = √7/4

tan(A) = Opposite side/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

BC/AB = 3k/√7k = 3/√7

tan A = 3/√7