91Ó°ÊÓ

We are given a figure

$$\begin{gathered} f\left(x\right)=\sin x \\ g\left(x\right)=\cos x,\text{and} \\ h\left(x\right)=\tan x \end{gathered}$$

It can be seen that $(x,1)$is a point on the graph of the function $x^2+y^2=4$. Substituting $1$ for y and solve the equation, we get

$$\begin{gathered} x^2+1^2=4 \\ {x}^2+1=4 \\ x=\sqrt{3} \end{gathered}$$

So, we get the side opposite to the angle $\mathrm{Î\pm }$as $1$, adjacent side as $\sqrt{3}$, and the hypotenuse as $2$.

So,

$$\begin{gathered} \cos \mathrm{Î\pm }=\frac{\sqrt{3}}{2} \\ \sin \mathrm{Î\pm }=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \tan \mathrm{Î\pm }=\frac{\sin \mathrm{Î\pm }}{\cos \mathrm{Î\pm }} \\ =\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \\ =\frac{1}{\sqrt{3}} \end{gathered}$$

Similarly, for second figure $\left(\frac{1}{3},y\right)$is a point on the graph of $x^2+y^2=1$.

Substituting $x=\frac{1}{3}$and solve the equation, we get

$$\begin{gathered} {\left(\frac{1}{3}\right)}^2+y^2=1 \\ \frac{1}{9}+y^2=1 \\ {y}^2=1-\frac{1}{9} \\ y=\sqrt{\frac{8}{9}} \\ y=Â\pm \frac{{\displaystyle 2\sqrt{2}}}{3} \\ y=-\frac{{\displaystyle 2\sqrt{2}}}{3} \end{gathered}$$

Because it is in the fourth quadrant.

From the figure, we have $\mathrm{θ}=2\mathrm{π}-\mathrm{β}$. So,

$$\begin{gathered} \sin \mathrm{θ}=\sin (2\mathrm{π}-\mathrm{β}) \\ \sin (2\mathrm{π}-\mathrm{β})=\sin (-\mathrm{β}) \\ =-\sin \mathrm{β} \end{gathered}$$

and

$$\begin{gathered} \cos \mathrm{θ}=\cos (2\mathrm{π}-\mathrm{β}) \\ \cos (2\mathrm{π}-\mathrm{β})=\cos (-\mathrm{β}) \\ =\cos \mathrm{β} \end{gathered}$$

So,

$$\begin{gathered} \sin \mathrm{β}=-\frac{2\sqrt{2}}{3} \\ \cos \mathrm{β}=\frac{1}{3} \end{gathered}$$

$$\begin{gathered} \tan \mathrm{β}=\frac{\sin \mathrm{β}}{\cos \mathrm{β}} \\ =\frac{\left(-\frac{2\sqrt{2}}{3}\right)}{\left(\frac{1}{3}\right)} \\ =-2\sqrt{2} \end{gathered}$$

The value of function is given by,

$$\begin{gathered} h(\mathrm{Î\pm }+\mathrm{β})=\tan (\mathrm{Î\pm }+\mathrm{β}) \\ =\frac{\tan \mathrm{Î\pm }+\tan \mathrm{β}}{1-\tan \mathrm{Î\pm }\tan \mathrm{β}} \\ =\frac{\frac{1}{\sqrt{3}}+(-2\sqrt{2})}{1-\left(\frac{1}{\sqrt{3}}\right)(-2\sqrt{2})} \\ =\frac{\left(\frac{1-2\sqrt{6}}{\sqrt{3}}\right)}{\left(\frac{\sqrt{3}+2\sqrt{2}}{\sqrt{3}}\right)} \\ =\frac{1-2\sqrt{6}}{\sqrt{3}+2\sqrt{2}} \end{gathered}$$

Now, multiplying and dividing by $\sqrt{3}-2\sqrt{2}$, we get

$$\begin{gathered} =\frac{1-2\sqrt{6}}{\sqrt{3}+2\sqrt{2}}×\frac{\sqrt{3}-2\sqrt{2}}{\sqrt{3}-2\sqrt{2}} \\ =\frac{(1-2\sqrt{6})(\sqrt{3}-2\sqrt{2})}{(\sqrt{3}{)}^2-(2\sqrt{2}{)}^2} \\ =\frac{\sqrt{3}-2\sqrt{2}-2\sqrt{18}+4\sqrt{12}}{3-8} \\ =\frac{\sqrt{3}-2\sqrt{2}-6\sqrt{2}+8\sqrt{3}}{-5} \\ =\frac{9\sqrt{3}-8\sqrt{2}}{-5} \\ =\frac{8\sqrt{2}-9\sqrt{3}}{5} \end{gathered}$$