If $ 4 \tan \theta=3 $, evaluate $ \frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1} $

Given:

\( 4 \tan \theta=3 \)

To do:

We have to evaluate \( \frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1} \).

Solution:  

$4 \tan \theta=3$

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

We know that,

$\sec ^{2} \theta=1+\tan ^{2} \theta$

$\cos ^{2} \theta=\frac{1}{\sec ^{2} \theta}$

$=\frac{1}{1+\tan ^{2} \theta}$
$\Rightarrow \cos \theta=\frac{1}{\sqrt{1+\tan ^{2} \theta}}$

$=\frac{1}{\sqrt{1+(\frac{3}{4})^2}}$

$=\frac{1}{\sqrt{1+\frac{9}{16}}}$

$=\frac{1}{\sqrt{\frac{16+9}{16}}}$

$=\frac{1}{\sqrt{\frac{25}{16}}}$

$=\frac{1}{\frac{5}{4}}$

$=\frac{4}{5}$

$\sin \theta=\sqrt{1-\cos^2 \theta}$

$=\sqrt{1-(\frac{4}{5})^2}$

$=\sqrt{\frac{25-16}{25}}$

$=\sqrt{\frac{9}{25}}$

$=\frac{3}{5}$

This implies,

$\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1}=\frac{4\left(\frac{3}{5}\right) -\frac{4}{5} +1}{4\left(\frac{3}{5}\right) +\frac{4}{5} -1}$

$=\frac{\frac{12}{5} -\frac{4}{5} +1}{\frac{12}{5} +\frac{4}{5} -1}$

$=\frac{\frac{12-4+5}{5}}{\frac{12+4-5}{5}}$

$=\frac{13}{11}$

The value of \( \frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1} \) is $\frac{13}{11}$.

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Updated on: 10-Oct-2022

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