Question

If $2\sin x=1,\frac{{\pi }^c}{2}<x<{\pi }^c$ and $\sqrt{2}\cos y=1,\frac{3{\pi }^c}{2}<y<2{\pi }^c$. find the value of $\frac{\tan x+\tan y}{\cos x-\cos y}$

Answer 1

$\begin{array}{c}2\sin x=1\\ \sin x=\frac{1}{2}\\ x={30}^{\circ }\\ \sqrt{2}\cos y=1\\ \cos y=\frac{1}{\sqrt{2}}\\ y={45}^{\circ }\\ thevalueof\frac{\tan x+\tan y}{\cos x-\cos y}=\frac{\tan 30+\tan 45}{\cos 30-\cos 45}\\ =\frac{\frac{1}{\sqrt{3}}+1}{\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}}\\ =\frac{\frac{1}{1.732}+1}{\frac{1.732}{2}-\frac{1}{1.414}}\\ =\frac{1.577}{.\theta 66-.707}=9.918\end{array}$