If $\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=\frac{3}{5}$, then the value of $\frac{{cosec} \theta+\cot \theta}{{cosec} \theta-\cot \theta}$ is :

$31+8 \sqrt{15}$

\(\frac{secθ - tanθ}{secθ + tanθ}\) = \(\frac{3}{5}\)

By using componendo and dividendo

\(\frac{secθ }{ tanθ}\) = \(\frac{3 +5}{5 - 3}\)

\(\frac{secθ }{ tanθ}\) = \(\frac{8}{2}\)

\(\frac{1 }{sinθ}\) = \(\frac{4}{1}\)

sinθ = \(\frac{1}{4}\)

{ sinθ = \(\frac{P}{H}\) }

By using pythagoras theorem,

P² + B² = H²

1² + B² = 4²

B = \(\sqrt {15 }\)

Now,

\(\frac{cosecθ+ cotθ}{cosecθ- cotθ}\)

= \(\frac{H+ B}{H- B}\)

= \(\frac{4+ √15}{4- √15}\)

= \(\frac{4+ √15}{4- √15}\) x \(\frac{4+ √15}{4+ √15}\)

= (4+ √15)²

= 16 + 15 + 8√15

= 31 + 8√15