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 $3$.   

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Updated on: 10-Oct-2022

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