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If $ \operatorname{cosec} \theta=\frac{13}{12} $, find the value of $ \frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta} $
Given:
\( \operatorname{cosec} \theta=\frac{13}{12} \)
To do:
We have to find the value of \( \frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta} \).
Solution:
We know that,
$\sin \theta=\frac{1}{\operatorname{cosec} \theta}$
$=\frac{1}{\frac{13}{12}}$
$=\frac{12}{13}$
$\Rightarrow \cos^{2} \theta=1-\sin^2 \theta$
$=1-(\frac{12}{13})^2$
$=1-\frac{144}{169}$
$=\frac{169-144}{169}$
$=\frac{25}{169}$
$\Rightarrow \cos \theta=\sqrt{\frac{25}{169}}$
$=\frac{5}{13}$
Therefore,
$\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}=\frac{2(\frac{12}{13})-3(\frac{5}{13})}{4(\frac{12}{13})-9(\frac{5}{13})}$
$=\frac{\frac{24-15}{13}}{\frac{48-45}{13}}$
$=\frac{9}{3}$
$=3$
The value of \( \frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta} \) is $3$.