If $\frac{cosec~\theta+\cot \theta}{cosec~\theta-\cot \theta}=7$, then the value of $\frac{4 \sin ^2 \theta-1}{4 \sin ^2 \theta+5}$ is:
Answer & explanation
Correct answer: option 4
We are given that :-
\(\frac{ cosecθ + cotθ}{cosecθ - cotθ }\) = 7
cosecθ + cotθ = 7cosecθ - 7cotθ
8 cotθ = 6 cosecθ
cosθ = \(\frac{ 3}{4}\)
Now, cos²θ = \(\frac{ 9}{16}\)
{ using , sin²θ + cos²θ = 1 }
1 - sin²θ = \(\frac{ 9}{16}\)
sin²θ = 1 - \(\frac{ 9}{16}\)
sin²θ = \(\frac{ 7}{16}\)
Now,
\(\frac{ 4sin²θ - 1 }{4sin²θ + 5 }\)
= \(\frac{ 4 × 7/16 - 1 }{4×7/16 + 5 }\)
= \(\frac{ 3/4 }{27/4 }\)
= \(\frac{ 1 }{9 }\)