Solution
$$\cos\ \theta\ =\frac{B}{H}=\frac{4}{5}$$
P= $$\sqrt{\ 5^2-4^2}=\sqrt{\ 9}=3$$
$$(\sec \theta + \cosec \theta)$$
$$\frac{H}{B}+\frac{H}{P}$$
$$\frac{5}{4}+\frac{5}{3}=\frac{35}{12}$$.
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