$\tan{(45^°)}$ Proof

The value of tan of $45$ degrees can be derived mathematically in three ways. One of them is a trigonometric approach and the other two methods are purely geometrical approaches but there is a slight difference between the geometrical methods.

Theoretical approach

The value of $\tan{(45^°)}$ can be derived exactly by theoretical approach of geometry on the basis of a geometric property, which reveals the direct relation between opposite and adjacent sides when angle of right angled triangle is $\dfrac{\pi}{4}$.

$\tan{(45^°)}$ $\,=\,$ $\dfrac{Length \, of \, Opposite \, side}{Length \, of \, Adjacent \, side}$

$\implies \tan{(45^°)} \,=\, \dfrac{QR}{PR}$

When angle of right triangle is $45$ degrees, the length of opposite side is exactly equal to the length of adjacent side. In this case, the length of both opposite side and adjacent side is denoted by $l$.

$\implies \tan{(45^°)} \,=\, \dfrac{l}{l}$

$\implies \tan{(45^°)} \,=\, \require{cancel} \dfrac{\cancel{l}}{\cancel{l}}$

$\,\,\, \therefore \,\,\,\,\,\, \tan{(45^°)} \,=\, 1$

Practical approach

It is not an issue if you don’t know the geometric property of opposite and adjacent sides when angle of right triangle is $50^g$. You can evaluate the exact value of $\tan{(50^g)}$ on your own by constructing a right triangle with $45$ degrees angle by your geometrical tools.

construction of right triangle with 45 degrees for tan 45 degrees value

Draw a straight line from a point $J$.
After that, coincide the point $J$ with centre of the protractor and also coincide the right side base line of protractor with horizontal line. Now, mark a point at $45$ degrees angle.
Draw a straight line by a ruler by Joining point $J$ and a point, marked at $45$ degrees angle.
Now, set some distance between needle and pencil points of compass. In this example, $12 \, cm$ is set. Then, draw an arc on the $45^°$ angle line from point $J$ and it cuts the line at point $K$.
Now, draw a perpendicular line to horizontal line from point $K$ by set square and it intersects the horizontal line at point $L$. It forms a right triangle geometrically, known as $\Delta LJK$.

Now, you can start the procedure to find the exact value of tan of angle $45$ degrees from $\Delta LJK$ by expressing it in terms of lengths of the respective sides.

$\tan{(45^°)} = \dfrac{Length \, of \, Opposite \, side}{Length \, of \, Adjacent \, side}$

$\implies \tan{(45^°)} \,=\, \dfrac{KL}{JL}$

Measuring opposite and adjacent sides of right triangle when angle is 45 degrees

In this example, the lengths of opposite and adjacent sides are unknown. So, measure the lengths of both opposite side ($KL$) and adjacent side ($JL$) by ruler.

You will observe that the lengths of both opposite side and adjacent side are equal and length of each of them is nearly $8.5$ centimetres.

$\implies \tan{(45^°)} \,=\, \dfrac{KL}{JL} = \dfrac{8.5}{8.5}$

$\implies \tan{(45^°)} \,=\, \require{cancel} \dfrac{\cancel{8.5}}{\cancel{8.5}}$

$\,\,\, \therefore \,\,\,\,\,\, \tan{(45^°)} \,=\, 1$

Trigonometric approach

The value of $\tan{\Big(\dfrac{\pi}{4}\Big)}$ can be evaluated exactly by the ratio or quotient identity of sin and cos functions. It is actually evaluated by the ratio of of sin $45$ degrees value to cos $45$ degrees value.

$\tan{(45^°)} \,=\, \dfrac{\sin{(45^°)}}{\cos{(45^°)}}$

$\implies \tan{(45^°)} \,=\, \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}}$

$\implies \tan{(45^°)} \,=\, \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{2}}{1}$

$\implies \tan{(45^°)} \,=\, \dfrac{1 \times \sqrt{2}}{\sqrt{2} \times 1}$

$\implies \tan{(45^°)} \,=\, \dfrac{\sqrt{2}}{\sqrt{2}}$

$\implies \tan{(45^°)} \,=\, \require{cancel} \dfrac{\cancel{\sqrt{2}}}{\cancel{\sqrt{2}}}$

$\,\,\, \therefore \,\,\,\,\,\, \tan{(45^°)} \,=\, 1$