If $\tan \theta=\sqrt{5}$, then the value of $\frac{{cosec}^2 \theta+\sec ^2 \theta}{{cosec}^2 \theta-\sec ^2 \theta}$ is:
Answer & explanation
Correct answer: option 3
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,
\(\frac{ cosec²θ + sec²θ }{cosec²θ - sec²θ }\)
= \(\frac{ ( H/P)² + ( H/B)² }{( H/P)² - ( H/B)² }\)
= \(\frac{ 6/51 + 6/1 }{6/5- 6/1 }\)
= \(\frac{ 36/5 }{ - 24/5 }\)
= - \(\frac{ 3 }{ 2 }\)