Question

If $\tan \theta =\frac{5}{12}$ and $\theta$ is not in the fourth quardrant then $\frac{\tan ({90}^{\circ }+\theta )-\sin ({180}^{\circ }-\theta )}{\sin ({270}^{\circ }-\theta )+cosec({360}^{\circ }-\theta )}=$

• A. $\frac{109}{131}$
• B. $-\frac{109}{131}$
• C. $\frac{131}{109}$
• D. $-\frac{131}{109}$

Answer 1

The correct option is D $-\frac{131}{109}$

$\tan \theta =\frac{5}{12}.\theta ϵII;G.E=\frac{-\cot \theta -\sin \theta }{-\cos \theta -cosec\theta }=\frac{\frac{12}{5}-\frac{5}{13}}{\frac{12}{13}-\frac{13}{5}}$