In a right -angled triangle  ABC, right – angled at B, if tanA = \frac{3}{4}, then find cosec A.

Given:
In right-angled triangle ABC, ∠B = 90° and
$\tan A = \frac{3}{4}$

We know that
$\tan A = \frac{\text{Opposite}}{\text{Adjacent}}$

So take
Opposite side = 3k
Adjacent side = 4k

Using the Pythagoras theorem:

Hypotenuse
$= \sqrt{(3k)^2 + (4k)^2}$
$= \sqrt{9k^2 + 16k^2}$
$= \sqrt{25k^2}$
$= 5k$

Now
$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3k}{5k} = \frac{3}{5}$

Therefore
$\cosec A = \frac{1}{\sin A} = \frac{5}{3}$

Final Answer: $\frac{5}{3}$