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:

$\frac{10}{49}$

$tan θ=\frac{2}{\sqrt{11}}, 0 < θ < 90°$

We know ,

tanθ = \(\frac{P }{B}\) = \frac{2}{\sqrt{11}}

tan²θ  = \(\frac{4 }{11}\)

H² = 4 + 11 = 15

H = \(\sqrt {15 }\)

$\frac{2cosec^2θ-3sec^2θ}{3cosec^2θ+4sec^2θ}$

= \(\frac{2 cosec²θ - 3 sec²θ  }{ 3 cosec²θ + 4 sec²θ}\)

= \(\frac{2 - 3 tan²θ  }{ 3 + 4 tan²θ}\)

= \(\frac{2 - 3 × 4/11  }{ 3 + 4 × 4/11}\)

= \(\frac{22 - 12  }{ 33 + 16 }\)

= \(\frac{10  }{ 49 }\)