Question : If $\cos \theta=\frac{12}{13}$, then the value of $\frac{\sin \theta(1-\tan \theta)}{\tan \theta(1+\operatorname{cosec} \theta)}$ is:
Option 1: $\frac{25}{78}$
Option 2: $\frac{35}{234}$
Option 3: $\frac{35}{108}$
Option 4: $\frac{25}{156}$
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Correct Answer: $\frac{35}{234}$
Solution :
Given:
$\cos \theta=\frac{12}{13}$
We know that,
$\cos \theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{12}{13}$
Using Pythagoras theorem, we get, Perpendicular = $\sqrt{13^2-12^2}=5$
$\sin \theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{5}{13}$
$\tan \theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{5}{12}$
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{13}{5}$
Now,
$\frac{\sin \theta(1-\tan \theta)}{\tan \theta(1+\operatorname{cosec} \theta)}$
$= \frac{\frac{5}{13}(1-\frac{5}{12})}{\frac{5}{12}(1+\frac{13}{5})}$
$= \frac{\frac{5}{13}\times \frac{7}{12}}{\frac{5}{12}\times \frac{18}{5}}$
$=\frac{35}{234}$
Hence, the correct answer is $\frac{35}{234}$.
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