In the right triangle shown in the figure ,what is the value of cosec $\theta$ ?

$\frac{13}{5}$ $\frac{5}{13}$ $\frac{5}{11}$ $\frac{12}{13}$

$\frac{13}{5}$

By using pythagoras theorem , P² + B² = H² ATQ, B = 5 and H = 13 P² + 5² = 13² P² = 169 - 25 P² = 144 P = 12 cosecθ = \(\frac{H}{P}\) = \(\frac{13}{5}\)