Question

If $\tan \theta =-5/12,\theta$ is not in the second quadrant, then show that
$\frac{\sin ({360}^o-\theta )+\tan ({90}^o+\theta )}{-sec({270}^o+\theta )+cosec(-\theta )}$ = $\frac{181}{338}$

Answer 1

Since $\tan$ is negative and not in 2nd quadrant, then it must be 4th quadrant.

$\tan \theta =\frac{-5}{12}$

$\sec \theta =\sqrt{1+{\tan }^2\theta }=\frac{13}{12}$ since $\sec$ is positive in 4th quadrant

$\cos \theta =\frac{1}{\sec \theta }=\frac{12}{13}$

$\sin \theta =-\sqrt{1-{\cos }^2\theta }=\frac{-5}{13}$ since $\sin$ is -ve in 4th quadrant

$⟹cosec\theta =\frac{1}{\sin \theta }=\frac{-13}{5}$

Now,

$\frac{\sin ({360}^o-\theta )+\tan ({90}^o+\theta )}{-\sec ({270}^o+\theta )+cosec(-\theta )}$

$=\frac{-\sin \theta -cot\theta }{-cosec\theta -cosec\theta }$

$=\frac{\frac{5}{13}+\frac{12}{5}}{\frac{13}{5}+\frac{13}{5}}$

$=\frac{25+156}{65}\times \frac{5}{26}=\frac{181\times 5}{65\times 26}$

$=\frac{181}{338}$