If $cosec \theta = 1.25,$ then $\frac{4 \tan \theta - 5 \cos \theta + 1}{\sec \theta + 4 \cot \theta -1} = ?$

$\frac{10}{11}$

cosecθ = 1.25

cosecθ = \(\frac{5}{4}\)

{  we know that ,cosecθ = \(\frac{H}{P}\) }

By using pythagoras theorem ,

P² + B² = H²

4² + B² = 5²

B² = 25 - 16

B² = 9

B = 3

Now,

\(\frac{4tanθ - 5cosθ + 1 }{secθ +4 cotθ - 1 }\)

= \(\frac{4 × 4/3 - 5 × 3/5  + 1 }{5/3 + 4×3/4  - 1 }\) 

= \(\frac{16/3 - 3  + 1 }{5/3 + 3 - 1 }\)

= \(\frac{10 }{11 }\)