If 5 cos θ = 4 sin θ, 0° ≤ θ ≤ 90°, then what will be the value of sec  θ ?

$\frac{\sqrt{41}}{5}$ $\frac{3}{5}$ $\frac{\sqrt{41}}{16}$ $\frac{\sqrt{41}}{4}$

$\frac{\sqrt{41}}{4}$

5 cos θ = 4 sin θ cot θ = \(\frac{4}{5}\)  { cot θ = \(\frac{B}{P}\) } By using pythagoras theorem, P² + B² = H² 5² + 4² = H² H = √41 Now, secθ  = \(\frac{H}{B}\) = \(\frac{√41 }{4}\)