If cosecθ = 1.25

find the value of \(\frac{4tanθ - 5cosθ+1}{secθ + 4cotθ-1}\)+\(\frac{10}{11}\).

\(\frac{20}{11}\)

cosecθ =\(\frac{125}{100}\)=\(\frac{5}{4}\)=\(\frac{H}{P}\) (Triplet 3, 4, 5)

B = 3

Put than in find

⇒ \(\frac{4×\frac{4}{3}-5×\frac{3}{5}+1}{\frac{5}{3}+4×\frac{3}{4}-1}\)+\(\frac{10}{11}\) ⇒ \(\frac{\frac{16}{6}-3+1}{\frac{5}{3}+3-4}\)+\(\frac{10}{11}\)

⇒ \(\frac{\frac{10}{3}}{\frac{11}{3}}\)+\(\frac{10}{11}\)

= \(\frac{20}{11}\)