Prove the following

Question:

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

(a) $\frac{2}{3}$

(b) $\frac{1}{3}$

(c) $\frac{1}{2}$

(d) $\frac{3}{4}$

Solution:

(c) Given, $4 \tan \theta=3$

$\Rightarrow$ $\tan \theta=\frac{3}{4}$  ...(i)

$\therefore$ $\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}=\frac{4 \frac{\sin \theta}{\cos \theta}-1}{4 \frac{\sin \theta}{\cos \theta}+1}$

[divide by $\cos \theta$ in both numerator and denominator]

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

$=\frac{4\left(\frac{3}{4}\right)-1}{4\left(\frac{3}{4}\right)+1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}$ [put the value from Eq. (i)]

 

 

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