If $sin θ = \frac{11}{15},$ then the value of $(sec θ - tan θ)$ is :

$\frac{2\sqrt{26}}{13}$ $\frac{\sqrt{26}}{13}$ $\frac{4}{\sqrt{26}}$ $\frac{1}{\sqrt{26}}$

$\frac{\sqrt{26}}{13}$

sin θ  = \(\frac{11}{15}\) { sin θ = \(\frac{P}{H}\) } By using pythagoras theorem , P² + B² = H² 11² + B² = 15² B² = 225 - 121 B = 2√26 Now, secθ - tanθ = \(\frac{H}{B}\) - \(\frac{P}{B}\) = \(\frac{15}{2√26 }\) - \(\frac{11}{2√26 }\) = \(\frac{4}{2√26 }\) = \(\frac{√26}{13 }\)