If $\tan \theta=\sqrt{5}$, then the value of $\frac{{cosec}^2 \theta+\sec ^2 \theta}{{cosec}^2 \theta-\sec ^2 \theta}$ is:

$-\frac{3}{2}$

We are given that :-

tan θ = \(\frac{ √5 }{1}\)

{ using , tan θ = \(\frac{ P }{B}\) }

By using pythagoras theorem,

P² + B² = H²

5 + 1 = H²

H = √6

Now,

\(\frac{ cosec²θ + sec²θ   }{cosec²θ  - sec²θ  }\)

= \(\frac{ ( H/P)² + ( H/B)²   }{( H/P)² - ( H/B)²  }\)

= \(\frac{ 6/51 + 6/1   }{6/5- 6/1  }\)

= \(\frac{ 36/5 }{ - 24/5  }\)

= - \(\frac{ 3 }{ 2 }\)