Question

If $4\tan \left(\mathrm{\theta }\right)=3$, then $\frac{4\sin \left(\theta \right)-\cos \left(\theta \right)}{4\sin \left(\theta \right)+\cos \left(\theta \right)}$ is equal to:

• A. $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• B. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• C. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• D. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Answer 1

The correct option is C

$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

We have:

$$\begin{gathered} 4\tan \left(\theta \right)=3 \\ \Rightarrow \tan \left(\theta \right)=\frac{3}{4}......\left(i\right) \end{gathered}$$

Now,

$\frac{4\sin \left(\theta \right)-\cos \left(\theta \right)}{4\sin \left(\theta \right)+\cos \left(\theta \right)}$ $=\frac{4{\displaystyle \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}}-{\displaystyle \frac{\cos \left(\theta \right)}{\cos \left(\theta \right)}}}{4{\displaystyle \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}}+{\displaystyle \frac{\cos \left(\theta \right)}{\cos \left(\theta \right)}}}\left[Dividingnumeratoranddenominatorby\cos \left(\theta \right)\right]$

$=\frac{4\tan \left(\theta \right)-{\displaystyle 1}}{4\tan \left(\theta \right)+{\displaystyle 1}}[\because \tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}]$

$$\begin{gathered} =\frac{4{\displaystyle x\frac{3}{4}}-{\displaystyle 1}}{4{\displaystyle x\frac{3}{4}}+1}[U\sin g(i\left)\right] \\ =\frac{2}{4} \\ =\frac{1}{2} \end{gathered}$$

Final answer: Hence, option C is correct.