91Ó°ÊÓ

The figure

The function $$\begin{gathered} f\left(x\right)=\sin x \\ g\left(x\right)=\cos x \\ h\left(x\right)=\tan x \end{gathered}$$

To evaluate the value of $h\left(2\mathrm{θ}\right)$

$$\begin{gathered} h\left(x\right)=\tan x \\ h\left(2\mathrm{θ}\right)=\tan 2\mathrm{θ} \end{gathered}$$

From the figure, the equation of circle is

$x^2+y^2=5$

Substitute $(a,2)$in the equation,

$$\begin{gathered} a^2+2^2=5 \\ {a}^2=5-4 \\ {a}^2=1 \\ a=Â\pm 1 \end{gathered}$$

The value of $a=-1$since the point is in second quadrant.

So the point is$(-1,2)$

The required angle is given by the difference between the given angle and the right angle angle formed by the coordinates.

So to find the angle $\mathrm{θ}-\frac{\mathrm{π}}{2}$, by using the Pythagorean theorem,

$\sqrt{1^2+2^2}=\sqrt{5}$

So,

$$\begin{gathered} \sin (\mathrm{θ}-\frac{\mathrm{π}}{2})=\frac{1}{\sqrt{5}} \\ \cos (\mathrm{θ}-\frac{\mathrm{π}}{2})=\frac{2}{\sqrt{5}} \end{gathered}$$

By odd-even identities,

$$\begin{gathered} \sin (-\mathrm{θ})=-\sin \mathrm{θ} \\ \sin (-(\frac{\mathrm{π}}{2}-\mathrm{θ}\left)\right)=-\sin (\frac{\mathrm{π}}{2}-\mathrm{θ}) \\ =-\frac{1}{\sqrt{5}} \\ \cos (-\mathrm{θ})=\cos \mathrm{θ} \\ \cos (-(\frac{\mathrm{π}}{2}-\mathrm{θ}\left)\right)=\cos (\frac{\mathrm{π}}{2}-\mathrm{θ}) \\ =\frac{2}{\sqrt{5}} \end{gathered}$$

By using the identity,

$$\begin{gathered} \sin (\frac{\mathrm{π}}{2}-\mathrm{θ})=\cos \mathrm{θ} \\ =-\frac{1}{\sqrt{5}} \\ \cos (\frac{\mathrm{π}}{2}-\mathrm{θ})=\sin \mathrm{θ} \\ =\frac{2}{\sqrt{5}} \end{gathered}$$

We know that,

$$\begin{gathered} \tan \mathrm{θ}=\frac{\sin \mathrm{θ}}{\cos \mathrm{θ}} \\ =\frac{{\displaystyle \frac{2}{\sqrt{5}}}}{{\displaystyle \frac{-1}{\sqrt{5}}}} \\ =-2 \end{gathered}$$

By double-angle formula,

$$\begin{gathered} \tan 2\mathrm{θ}=\frac{2\tan \mathrm{θ}}{1-{\tan }^2\mathrm{θ}} \\ =\frac{2(-2)}{1-{(-2)}^2} \\ =\frac{-4}{1-4} \\ =\frac{4}{3} \end{gathered}$$