(a) \(\frac7{23}\)
8 tan A = 15, ⇒ tan A = \(\frac{15}{8}\)
Now, \(\frac{sin\,A-cos\,A}{sin\,A+cos\,A}\) = \(\frac{\frac{sin\,A}{cos\,A}-\frac{cos\,A}{cos\,A}}{\frac{Sin\,A}{cos\,A}+\frac{cos\,A}{cos\,A}}\)
= \(\frac{tan\,A-1}{tan\,A+1}\)
Now, substitute the value of tan A.