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in Trigonometry by (49.2k points)

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) \(\frac{17}{23}\)

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1 Answer

+1 vote
by (46.2k points)

(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.

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