Question

Question 12
If $4tan\theta =3$, then $\left(\frac{4sin\theta -cos\theta }{4sin\theta +cos\theta }\right)$ is equal to

(A) $\frac{2}{3}$
(B) $\frac{1}{3}$
(C) $\frac{1}{2}$
(D) $\frac{3}{4}$

Answer 1

The answer is (C).

Given, $4tan\theta =3$

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

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

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

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

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