If $\frac{\sin A+\cos A}{\cos A}=\frac{17}{12}$, then the value of $\frac{1-\cos A}{\sin A}$ is:

$\frac{1}{5}$

\(\frac{sinA + cosA }{cosA}\) = \(\frac{17}{12}\)

12sinA  +  12cosA = 17 cosA

12sinA = 5cosA

tanA = \(\frac{5}{12}\)

{ we know, tanA = \(\frac{P}{B}\) }

By using pythagoras theorem ,

P² + B² = H²

5² + 12² = H²

H = 13

Now,

\(\frac{1-cosA}{sinA}\)

= \(\frac{1-cosA}{sinA}\)

= \(\frac{1-B/H}{P/H}\)

= \(\frac{H - B}{P}\)

= \(\frac{13 - 12}{5}\)

= \(\frac{1}{5}\)