If $ 8 \tan A=15, $ then the value of $ \frac{\sin A-\cos A}{\sin A+\cos A} $ is
(a) $ \frac{7}{23} $
(b) $ \frac{11}{23} $
(c) $ \frac{13}{23} $
(d) $ \frac{17}{23} $
Given:
\( 8 \tan A=15 \)
To do:
We have to find the value of \( \frac{\sin A-\cos A}{\sin A+\cos A} \).
Solution:
We know that,
$tan\ A=\frac{opposite}{adjacent}$ and
\( (Hypotenuse)^2=(opposite)^2+(adjacent)^2 \)
\( 8 \tan A=15 \)
$tan\ A=\frac{15}{8}$
Opposite$=15$ and Adjacent$=8$
\( (Hypotenuse)^2=(15)^2+(8)^2 \)
Hyotenuse$=\sqrt{225+64}$
$=\sqrt{289}$
$=17$
$sin\ A=\frac{opposite}{Hypotenuse}$
$=\frac{15}{17}$
$cos\ A=\frac{adjacent}{Hypotenuse}$
$=\frac{8}{17}$
Therefore,
$\frac{sin\ A-cos\ A}{sin\ A+cos\ A}=\frac{\frac{15}{17}-\frac{8}{17}}{\frac{15}{17}+\frac{8}{17}}$
$=\frac{\frac{15-8}{17}}{\frac{15+8}{17}}$
$=\frac{\frac{7}{17}}{\frac{23}{17}}$
$=\frac{7}{23}$
Option (a) is the correct answer.
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