Question

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

Answer 1

$tan\theta =\left(\frac{-5}{12}\right)=-veand\theta isnotinsecondquadrantthen\theta isinforthquadrantandhencesin\theta =\left(\frac{-5}{13}\right)andcos\theta =\left(\frac{12}{13}\right)now\left(\frac{sin({360}^0-\theta )-cot\theta }{-sec({270}^0+\theta )+cosec(-\theta )}\right)=\left(\frac{sin(-\theta )-cot\theta }{-cosec\theta -cosec\theta }\right)=\left(\frac{-sin\theta -cot\theta }{-2cosec\theta }\right)=\left(\frac{sin\theta +cot\theta }{2cosec\theta }\right)=\left(\frac{\left(\frac{-5}{13}\right)-\left(\frac{12}{5}\right)}{2\times \left(\frac{-13}{5}\right)}\right)=\left(\frac{25+156}{13\times 5\times 2\times 13}\right)\times 5=\left(\frac{181}{338}\right)=RHS$