If $\frac{\sin A+\cos A}{\cos A}=\frac{17}{12}$, then the value of $\frac{1-\cos A}{\sin A}$ is:
Answer & explanation
Correct answer: option 4
\(\frac{sinA + cosA }{cosA}\) = \(\frac{17}{12}\)
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,
\(\frac{1-cosA}{sinA}\)
= \(\frac{1-cosA}{sinA}\)
= \(\frac{1-B/H}{P/H}\)
= \(\frac{H - B}{P}\)
= \(\frac{13 - 12}{5}\)
= \(\frac{1}{5}\)