Question

If $\sin \alpha =\frac{5}{13}$ and $\frac{\pi }{2}<\alpha <\pi$, then the value of $\frac{\sec \alpha +\tan \alpha }{cosec\alpha -\cot \alpha }$ is

• A. $0.3$
• B. $0.5$
• C. $-0.3$
• D. $-0.4$

Answer 1

The correct option is C $-0.3$

Given

$\frac{\pi }{2}<\alpha <\pi$

$⟹\cos \alpha <0\tan \alpha <0$

$⟹\sin \alpha =\frac{5}{13}⟹\csc \alpha =\frac{13}{5}$

$\cos \alpha =\sqrt{1-{\sin }^2\alpha }$

$⟹\cos \alpha =\sqrt{1-{\left(\frac{5}{13}\right)}^2}$

$⟹\cos \alpha =\sqrt{{\left(\frac{12}{13}\right)}^2}$

$⟹\cos \alpha =\frac{-12}{13}$

$(\cos \alpha <0)$

$⟹\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$

$⟹\tan \alpha =\frac{\frac{5}{13}}{\frac{-12}{13}}$

$⟹\tan \alpha =\frac{-5}{12}$

$⟹\cot \alpha =\frac{-12}{5}$

$⟹\sec \alpha =\frac{-13}{12}$

$\frac{\sec \alpha +\tan \alpha }{\csc \alpha -\cot \alpha }$

$⟹\frac{\frac{-13}{12}+\frac{-5}{12}}{\frac{13}{5}-\left(\frac{-12}{5}\right)}$

$⟹\frac{\frac{-18}{12}}{\frac{25}{5}}$

$⟹\frac{-3}{2}\times \frac{1}{5}$

$⟹\frac{-3}{10}=-0.3$