If $tan θ=\frac{2}{\sqrt{11}}, 0 < θ < 90°$, then the value of $\frac{2cosec^2θ-3sec^2θ}{3cosec^2θ+4sec^2θ}$ is equal to:
Answer & explanation
Correct answer: option 4
$tan θ=\frac{2}{\sqrt{11}}, 0 < θ < 90°$
We know ,
tanθ = \(\frac{P }{B}\) = \frac{2}{\sqrt{11}}
tan²θ = \(\frac{4 }{11}\)
H² = 4 + 11 = 15
H = \(\sqrt {15 }\)
$\frac{2cosec^2θ-3sec^2θ}{3cosec^2θ+4sec^2θ}$
= \(\frac{2 cosec²θ - 3 sec²θ }{ 3 cosec²θ + 4 sec²θ}\)
= \(\frac{2 - 3 tan²θ }{ 3 + 4 tan²θ}\)
= \(\frac{2 - 3 × 4/11 }{ 3 + 4 × 4/11}\)
= \(\frac{22 - 12 }{ 33 + 16 }\)
= \(\frac{10 }{ 49 }\)