91Ó°ÊÓ

We are given,

$$\begin{gathered} \sin \mathrm{Î\pm }=\frac{3}{5},0<\mathrm{Î\pm }<\frac{\mathrm{Ï€}}{2} \\ \cos \mathrm{β}=\frac{2\sqrt{5}}{5},-\frac{\mathrm{Ï€}}{2}<\mathrm{β}<0 \end{gathered}$$

By using ${\sin }^2\mathrm{Î\pm }+{\cos }^2\mathrm{Î\pm }=1$, we get

$$\begin{gathered} {\cos }^2\mathrm{Î\pm }=1-{\sin }^2\mathrm{Î\pm } \\ \cos \mathrm{Î\pm }=\sqrt{1-{\sin }^2\mathrm{Î\pm }} \end{gathered}$$

Now, putting the value of $\sin \mathrm{Î\pm }$, we get

$$\begin{gathered} \cos \mathrm{Î\pm }=\sqrt{1-\frac{9}{25}} \\ \cos \mathrm{Î\pm }=\sqrt{\frac{25-9}{25}} \\ \cos \mathrm{Î\pm }=\sqrt{\frac{16}{25}} \\ \cos \mathrm{Î\pm }=\frac{4}{5} \end{gathered}$$

Similarly, $\cos \mathrm{β}=\frac{2\sqrt{5}}{5}$and $\mathrm{β}$lies in the interval role="math" localid="1646560106929" $\left[-\frac{\mathrm{π}}{2},0\right]$, that means $\sin \mathrm{β}<0$in the fourth quadrant.

Now, using ${\sin }^2\mathrm{β}+{\cos }^2\mathrm{β}=1$, we get

role="math" localid="1646560342971" $$\begin{gathered} \sin \mathrm{β}=-\sqrt{1-\frac{20}{25}} \\ \sin \mathrm{β}=-\sqrt{\frac{25-20}{25}} \\ \sin \mathrm{β}=-\sqrt{\frac{5}{25}} \\ \sin \mathrm{β}=-\frac{\sqrt{5}}{5} \end{gathered}$$

Now, using $\sin (\mathrm{Î\pm }+\mathrm{β})=\sin \mathrm{Î\pm }·\cos \mathrm{β}+\cos \mathrm{Î\pm }·\sin \mathrm{β}$and putting the values, we get

localid="1646561187724" $$\begin{gathered} \sin (\mathrm{Î\pm }+\mathrm{β})=\frac{3}{5}×\frac{2\sqrt{5}}{5}+\frac{4}{5}×-\frac{\sqrt{5}}{5} \\ =\frac{6\sqrt{5}}{25}-\frac{4\sqrt{5}}{25} \\ =\frac{2\sqrt{5}}{25} \end{gathered}$$

Using $\cos (\mathrm{Î\pm }+\mathrm{β})=\cos \mathrm{Î\pm }·\cos \mathrm{β}-\sin \mathrm{Î\pm }·\sin \mathrm{β}$, and putting the values, we get

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

Using $\sin (\mathrm{Î\pm }-\mathrm{β})=\sin \mathrm{Î\pm }·\cos \mathrm{β}-\cos \mathrm{Î\pm }·\sin \mathrm{β}$, and putting the values, we get

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

Using $\tan \mathrm{θ}=\frac{\sin \mathrm{θ}}{\cos \mathrm{θ}}$, and putting the values, we get

$$\begin{gathered} \tan \mathrm{Î\pm }=\frac{3/5}{4/5} \\ \tan \mathrm{Î\pm }=\frac{3}{5}·\frac{5}{4} \\ \tan \mathrm{Î\pm }=\frac{3}{4} \end{gathered}$$

and

role="math" localid="1646561852467" $$\begin{gathered} \tan \mathrm{β}=\frac{-\sqrt{5}/5}{2\sqrt{5}/5} \\ \tan \mathrm{β}=\frac{-\sqrt{5}}{5}×\frac{5}{2\sqrt{5}} \\ \tan \mathrm{β}=-\frac{1}{2} \end{gathered}$$

Using $\tan (\mathrm{Î\pm }-\mathrm{β})=\frac{\tan \mathrm{Î\pm }-\tan \mathrm{β}}{1+\tan \mathrm{Î\pm }·\tan \mathrm{β}}$, and putting the values, we get

role="math" localid="1646562172347" $$\begin{gathered} \tan (\mathrm{Î\pm }-\mathrm{β})=\frac{\frac{3}{4}+\frac{1}{2}}{1+{\displaystyle \frac{3}{4}}×{\displaystyle \frac{-1}{2}}} \\ =\frac{{\displaystyle \frac{5}{4}}}{{\displaystyle \frac{5}{8}}} \\ =\frac{5}{4}·\frac{8}{5} \\ =2 \end{gathered}$$