Question

The distance between the points (tan alpha,1), (0,2)​

Answer 1

Here:

$x_{1} =\tan \alpha, y_{1} =1$ and $x_{2} =0, y_{2} =2$

We have to find the distance between the given points.

solution:

We know that, the distance between two points

$=\sqrt{(x_{2}-x_{1} )^2+(x_{2}-y_{1} )^2}$

$=\sqrt{(0-\tan \alpha )^2+(2-1)^2}$

$=\sqrt{(-\tan \alpha )^2+(1)^2}$

$=\sqrt{\tan^2 \alpha +1}$

Using the trigonometric identity:

$\sec^2 A=\tan^2 A +1$

$=\sqrt{\sec^2 \alpha }$

= $\sec \alpha$

Hence, the distance between the given point is "$\sec \alpha$".

Answer 2

Distance formula:

If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are two points in Cartesian plane, then the distance between them is obtained by

$d=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$

Step-by-step explanation:

The given points are $(tan\alpha,1)$ and $(0,2)$.

Then the distance between these points is

$=\sqrt{(tan\alpha-0)^{2}+(1-2)^{2}}$ units

$=\sqrt{tan^{2}\alpha+1}$ units

$=\sqrt{sec^{2}\alpha}$ units,

since $sec^{2}\theta-tan^{2}\theta=1$

$=sec\alpha$ units

Answer:

The required distance is $sec\alpha$ units.