If 4 tan 0 = 3, then \(\left[\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta+1}\right]\) is : (1)
A. \(\frac{13}{5}\)
B. \(\frac{5}{21}\)
C. \(\frac{13}{21}\)
D. \(\frac{12}{13}\)
Answer:C
If 4 tan 0 = 3, then \(\left[\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta+1}\right]\) is : (1)
A. \(\frac{13}{5}\) B. \(\frac{5}{21}\) C. \(\frac{13}{21}\) D. \(\frac{12}{13}\)