If 8 tan A = 15, then the value of \(\frac{sin\,A-cos\,A}{sin\,A+cos\

1.	If 8 tan A = 15, then the value of \(\frac{sin\,A-cos\,A}{sin\,A+cos\,A}\) is(a) \(\frac7{23}\) (b) \(\frac{11}{23}\)(c) \(\frac{13}{23}\)(d) \(\frac{17}{23}\)
Answer» (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.

If 8 tan A = 15, then the value of \(\frac{sin\,A-cos\,A}{sin\,A+cos\,A}\) is(a) \(\frac7{23}\) (b) \(\frac{11}{23}\)(c) \(\frac{13}{23}\)(d) \(\frac{17}{23}\)

Answer»

(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.