$3 - c o t^{2} A + c o s e c^{2} A$

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Solution

$3 cos A = 4 sin A$
$tan A = \frac{3}{4}$
$tan A = \frac{P}{B} = \frac{3}{4}$
Now, Using Pythagoras Theorem,
$H^{2} = P^{2} + B^{2}$
$H^{2} = 3^{2} + 4^{2}$
$H = 5$
Now,
$3 - {cot}^{2} A + c o s e c^{2} A = 3 - (\frac{B}{P})^{2} + (\frac{H}{P})^{2}$
= $3 - (\frac{4}{3})^{2} + (\frac{5}{3})^{2}$
= $3 - \frac{16}{9} + \frac{25}{9}$
= $\frac{27 - 16 + 25}{9}$
= $4$

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