Question : If $4 \tan \theta=3$, then $\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1}=?$
Option 1: $\frac{14}{11}$
Option 2: $\frac{12}{11}$
Option 3: $\frac{10}{11}$
Option 4: $\frac{13}{11}$
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Correct Answer: $\frac{13}{11}$
Solution :
Let $\angle BCA =\theta$
$4 \tan \theta=3$
⇒ $\tan \theta = \frac{3}{4}$
So, let the height and base be 3 units and 4 units respectively.
⇒ $\sin \theta = \frac{3}{\sqrt{3^2+4^2}} = \frac{3}{5}$
⇒ $\cos \theta = \frac{4}{\sqrt{3^2+4^2}} = \frac{4}{5}$
So, $\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1}$
$=\frac{4 ×\frac{3}{5}-\frac{4}{5}+1}{4 ×\frac{3}{5}+\frac{4}{5}-1}$
$=\frac{\frac{12}{5}-\frac{4}{5}+1}{\frac{12}{5}+\frac{4}{5}-1}$
$=\frac{\frac{8}{5}+1}{\frac{16}{5}-1}$
$=\frac{\frac{13}{5}}{\frac{11}{5}}$
$=\frac{13}{11}$
Hence, the correct answer is $\frac{13}{11}$.
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