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

Given: $4tan\theta=3$

To do: To $evaluate ( \frac{4sin\theta-cos\theta+1}{ 4sin\theta+cos\theta-1})$

Solution:
Given that, $4tan\theta=3$

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

$tan^{2}\theta=( \frac{3}{4})^{2}=\frac{9}{16}$

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

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

                                            $=\frac{25}{16}$


$\therefore cos^{2}\theta=\frac{1}{sec^{2}\theta}=\frac{16}{25}$



$\Rightarrow cos\theta=\sqrt{\frac{16}{25}}=\frac{4}{5}$


As known, $sin^{2}\theta=1-cos^{2}\theta$


$\Rightarrow sin^{2}\theta=1-\frac{16}{25}=\frac{9}{25}$


$\Rightarrow sin\theta=\frac{3}{5}$


Now,  $( \frac{4sin\theta-cos\theta+1}{4sin\theta+cos\theta-1})$


$=( \frac{4×\frac{3}{5}-\frac{4}{5}+1}{4×\frac{3}{5}+\frac{4}{5}+1})$


$=( \frac{\frac{8}{5}+1}{\frac{16}{5}+1})$


$=( \frac{\frac{13}{5}}{\frac{21}{5}})$


$=\frac{13}{21}$
 

Updated on: 10-Oct-2022

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