Question

If $4\tan \theta =3$, evaluate $\left(\frac{4\sin \theta -\cos \theta +1}{4\sin \theta +\cos \theta -1}\right)$.

Answer 1

Given That

$4\tan \theta =3$

Or $\tan \theta =\frac{3}{4}$

Squaring both sides

${\tan }^2\theta =\frac{9}{16}$

$\Rightarrow {\sec }^2\theta -1=\frac{9}{16}$

$\Rightarrow {\sec }^2\theta =1+\frac{9}{16}$

$\Rightarrow {\sec }^2\theta =\frac{25}{16}$

$\Rightarrow \sec \theta =\frac{5}{4}$ (assuming $\theta$ is an acute angle)

Now for given expression

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

divide Numerator and Denominator by $cos\theta$ we get ,

$E=\frac{4\tan \theta -1+\sec \theta }{4\tan \theta +1-\sec \theta }$

Now substitute the values

$E=\frac{4\times \frac{3}{4}-1+\frac{5}{4}}{4\times \frac{3}{4}+1-\frac{5}{4}}$

$E=\frac{2+\frac{5}{4}}{4-\frac{5}{4}}$

$E=\frac{13}{11}$