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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, $( \frac{4sin\theta-cos\theta+1}{4sin\theta+cos\theta-1})$
$=( \frac{4×\frac{3}{5}-\frac{4}{5}+1}{4×\frac{3}{5}+\frac{4}{5}+1})$
$=( \frac{\frac{8}{5}+1}{\frac{16}{5}+1})$
$=( \frac{\frac{13}{5}}{\frac{21}{5}})$
$=\frac{13}{21}$
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