If $5 \sin \theta=4$, then the value of $\frac{\sec \theta+4 \cot \theta}{4 \tan \theta-5 \cos \theta}$ is:

2 $\frac{3}{2}$ $\frac{5}{4}$ 1

2

sinθ = \(\frac{4}{5} By using pythagoras theorem , P² + B² = H² 4² + B² = 5² B² = 25 - 16 B² = 9 B = 3 Now, \(\frac{ secθ + 4cotθ }{4tanθ - 5cosθ}\) = \(\frac{ 5/3  + 4 × 3/4 }{4× 4/3 - 5×3/5}\) =  \(\frac{ 5/3  + 3 }{16/3 - 3}\) =  \(\frac{ 14 }{7}\) = 2