91Ó°ÊÓ

We are given$\cos \mathrm{Î\pm }=\frac{\sqrt{5}}{5},0<\mathrm{Î\pm }<\frac{\mathrm{Ï€}}{2};\sin \mathrm{β}=-\frac{4}{5},-\frac{\mathrm{Ï€}}{2}<\mathrm{β}<0$

We have,

$$\begin{gathered} \sin \mathrm{Î\pm }=\sqrt{1-{\cos }^2\mathrm{Î\pm }} \\ \sin \mathrm{Î\pm }=\sqrt{1-{\left(\frac{\sqrt{5}}{5}\right)}^2} \\ \sin \mathrm{Î\pm }=\sqrt{1-\frac{5}{25}} \\ \sin \mathrm{Î\pm }=\sqrt{\frac{20}{25}} \\ \sin \mathrm{Î\pm }=\frac{2\sqrt{5}}{5} \end{gathered}$$

We have,

$$\begin{gathered} \cos \mathrm{β}=\sqrt{1-{\sin }^2\mathrm{β}} \\ \cos \mathrm{β}=\sqrt{1-{(-\frac{4}{5})}^2} \\ \cos \mathrm{β}=\sqrt{\frac{9}{25}} \\ \cos \mathrm{β}=\frac{3}{5} \end{gathered}$$

We have,

$$\begin{gathered} \sin (\mathrm{Î\pm }+\mathrm{β})=\sin \mathrm{Î\pm }\cos \mathrm{β}+\cos \mathrm{Î\pm }\sin \mathrm{β} \\ \sin (\mathrm{Î\pm }+\mathrm{β})=\left(\frac{2\sqrt{5}}{5}\right)\left(\frac{3}{5}\right)+\left(\frac{\sqrt{5}}{5}\right)(-\frac{4}{5}) \\ \sin (\mathrm{Î\pm }+\mathrm{β})=\frac{6\sqrt{5}}{5}-\frac{4\sqrt{5}}{5} \\ \sin (\mathrm{Î\pm }+\mathrm{β})=\frac{2\sqrt{5}}{25} \end{gathered}$$

We get,

$$\begin{gathered} \mathrm{co}s(\mathrm{Î\pm }+\mathrm{β})=\cos \mathrm{Î\pm }\cos \mathrm{β}-\sin \mathrm{Î\pm }\sin \mathrm{β} \\ \mathrm{co}s(\mathrm{Î\pm }+\mathrm{β})=\left(\frac{\sqrt{5}}{5}\right)\left(\frac{3}{5}\right)-\left(\frac{2\sqrt{5}}{5}\right)(-\frac{4}{5}) \\ \mathrm{co}s(\mathrm{Î\pm }+\mathrm{β})=\frac{3\sqrt{5}}{25}+\frac{8\sqrt{5}}{25} \\ \mathrm{co}s(\mathrm{Î\pm }+\mathrm{β})=\frac{11\sqrt{5}}{25} \end{gathered}$$

We have,

$$\begin{gathered} \sin (\mathrm{Î\pm }-\mathrm{β})=\sin \mathrm{Î\pm }\cos \mathrm{β}-\cos \mathrm{Î\pm }\sin \mathrm{β} \\ \sin (\mathrm{Î\pm }-\mathrm{β})=\left(\frac{2\sqrt{5}}{5}\right)\left(\frac{3}{5}\right)-\left(\frac{\sqrt{5}}{5}\right)(-\frac{4}{5}) \\ \sin (\mathrm{Î\pm }-\mathrm{β})=\frac{6\sqrt{5}}{25}+\frac{4\sqrt{5}}{25} \\ \sin (\mathrm{Î\pm }-\mathrm{β})=\frac{10\sqrt{25}}{25} \end{gathered}$$

We have,

$$\begin{gathered} \tan \mathrm{Î\pm }=\frac{\sin \mathrm{Î\pm }}{\cos \mathrm{Î\pm }} \\ \tan \mathrm{Î\pm }=\frac{{\displaystyle \frac{2\sqrt{5}}{5}}}{{\displaystyle \frac{\sqrt{5}}{5}}} \\ \tan \mathrm{Î\pm }=2 \end{gathered}$$

similarly

$$\begin{gathered} \tan \mathrm{β}=\frac{\sin \mathrm{β}}{\cos \mathrm{β}} \\ \tan \mathrm{β}=\frac{-{\displaystyle \frac{4}{5}}}{{\displaystyle \frac{3}{5}}} \\ \tan \mathrm{β}=-\frac{4}{3} \end{gathered}$$

Now

$$\begin{gathered} \tan (\mathrm{Î\pm }-\mathrm{β})=\frac{\tan \mathrm{Î\pm }-\tan \mathrm{β}}{1+\tan \mathrm{Î\pm }\tan \mathrm{β}} \\ \tan (\mathrm{Î\pm }-\mathrm{β})=\frac{2-(-{\displaystyle \frac{4}{3}})}{1+2(-{\displaystyle \frac{4}{3}})} \\ \tan (\mathrm{Î\pm }-\mathrm{β})=\frac{{\displaystyle \frac{6}{3}+\frac{4}{3}}}{1-{\displaystyle \frac{8}{3}}} \\ \tan (\mathrm{Î\pm }-\mathrm{β})=-2 \end{gathered}$$

The values are

$$\begin{gathered} a\left)\sin \right(\mathrm{Î\pm }+\mathrm{β})=\frac{2\sqrt{5}}{25} \\ b)\cos (\mathrm{Î\pm }+\mathrm{β})=\frac{11\sqrt{5}}{25} \\ c\left)\sin \right(\mathrm{Î\pm }-\mathrm{β})=\frac{10\sqrt{5}}{25} \end{gathered}$$