If $\frac{cosecθ+cotθ}{cosecθ-cotθ}=7,$ then the value of $\frac{4sin^2θ+5}{4sin^2θ-1}$ is :
Answer & explanation
Correct answer: option 3
\(\frac{cosecθ+ cotθ}{cosecθ - cotθ}\) = 7
\(\frac{1+ cosθ}{1 - cosθ}\) = 7
1+ cosθ = 7 - 7cosθ
8cosθ = 6
cosθ = \(\frac{3}{4}\)
{ we know, cosθ = \(\frac{B}{H}\) }
By using pythagoras theorem,
P² + B² = H²
P² + 3² = 4²
P = \(\sqrt {7 }\)
Now,
\(\frac{4 sin²θ + 5 }{4sin²θ - 1 }\)
= \(\frac{4 × 7/16 + 5 }{4× 7/16 - 1 }\)
= 9