Question

$\mathrm{If}sec\alpha =\frac{2}{\sqrt{3}},\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{1-\cos ec\alpha }{1+\cos ec\alpha },\mathrm{where}\mathrm{\alpha }\mathrm{is}\mathrm{in}\mathrm{IV}\mathrm{quadrant}.$

Answer 1

$$\begin{gathered} \mathrm{Since}\sec \alpha =\frac{2}{\sqrt{3}},\alpha \mathrm{lies}\mathrm{in}\mathrm{quadrant}\mathrm{IV}. \\ \therefore \cos \alpha =\frac{1}{sec\alpha }=\frac{1}{{\displaystyle \frac{2}{\sqrt{3}}}} \\ \Rightarrow \cos \alpha =\frac{\sqrt{3}}{2} \\ \mathrm{We}\mathrm{know}\mathrm{that}{\sin }^2\alpha +{\cos }^2\alpha =1 \\ \Rightarrow {\sin }^2\alpha +{\left(\frac{\sqrt{3}}{2}\right)}^2=1 \\ \Rightarrow \sin \alpha =\pm \sqrt{\frac{1}{4}}=-\frac{1}{2}(\because \alpha \mathrm{lies}\mathrm{in}\mathrm{quadrant}\mathrm{IV}) \\ \mathrm{Now},\cos ec\alpha =\frac{1}{\sin \alpha }=\frac{1}{-{\displaystyle \frac{1}{2}}}=-2 \\ \therefore \frac{1-\cos ec\alpha }{1+\cos ec\alpha }=\frac{1-(-2)}{1+(-2)}=-\frac{3}{1}=-3 \end{gathered}$$