if 6sec θ = 10, then find the value of (5cosecθ-3cotθ)/(4cosθ+3sinθ) .

6sec θ = 10

sec θ = \(\frac{ 5 }{3}\)

By using pythagoras theorem ,

P² + B² = H²

P² + 3² = 5²

P = 4

Now,

\(\frac{ 5cosecθ - 3cotθ }{4cosθ + 3sinθ}\)

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

= \(\frac{16/4 }{24/5}\)

= \(\frac{5 }{6}\)