Solveeit Logo

Question

Question: How do you find the exact value of \(\tan \dfrac{\pi }{4}\)?...

How do you find the exact value of tanπ4\tan \dfrac{\pi }{4}?

Explanation

Solution

Start with assuming an isosceles right-angled triangle ΔMNP\Delta MNP and with right angle at N and sides MN and NP equal to each other. This gives us NMP=NPM=45\angle NMP = \angle NPM = 45^\circ . Now use the definition of the tangent function, find the expression for it. Now substitute the values in it and simplify it to get the required value of tanπ4\tan \dfrac{\pi }{4} .

Complete step-by-step answer:
Here in this question, we are given an expression in tangent function, i.e. tanπ4\tan \dfrac{\pi }{4} and we have to find the exact value for this expression.
Before starting with the solution, we must understand a few concepts about tangent functions. Tan function (or tangent function) in a triangle is the ratio of the opposite side to that of the adjacent side. The tangent function is one of the three main primary trigonometric functions. In a right-triangle, tan is defined as the ratio of the length of the perpendicular side to that of the adjacent side i.e. the base.
As we know that an angle of π\pi radians is equal to 180180^\circ angle.
tanπ4=tan(1804)=tan45\Rightarrow \tan \dfrac{\pi }{4} = \tan \left( {\dfrac{{180^\circ }}{4}} \right) = \tan 45^\circ
For finding this value, we first take an isosceles right-angle triangle ΔMNP\Delta MNP with sides of length ‘m’ units. In this triangle, we have two equal sides, i.e. MN=NP=m unitsMN = NP = m{\text{ units}} . Since the two adjacent sides are equal in a triangle, therefore the angles opposite to these sides are also equal. Thus we have:
MN=NP=m\Rightarrow MN = NP = m and PMN=MPN=MNP2=902=45\angle PMN = \angle MPN = \dfrac{{\angle MNP}}{2} = \dfrac{{90^\circ }}{2} = 45^\circ
This information can be represented in a diagram as:

We can use the Pythagoras theorem, which states that the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. By this theorem, in the triangle ΔMNP\Delta MNP , we get:
PM2=MN2+NP2\Rightarrow P{M^2} = M{N^2} + N{P^2}
Now let’s substitute the known values in this equation:
PM2=MN2+NP2PM2=m2+m2\Rightarrow P{M^2} = M{N^2} + N{P^2} \Rightarrow P{M^2} = {m^2} + {m^2}
This equation can be easily solved to find the unknown length of hypotenuse PM
PM2=m2+m2PM2=2m2\Rightarrow P{M^2} = {m^2} + {m^2} \Rightarrow P{M^2} = 2{m^2}
Taking square root on both the sides, we have:
PM2=2m2PM=2m\Rightarrow \sqrt {P{M^2}} = \sqrt {2{m^2}} \Rightarrow PM = \sqrt 2 m
Therefore, we get the length of the hypotenuse PM as 2m\sqrt 2 m
Now, according to the definition of the tangent function, we have the relation:
tanθ=PerpendicularBase\Rightarrow \tan \theta = \dfrac{{Perpendicular}}{{Base}}
For the triangle ΔMNP\Delta MNP, using the tangent function in angle PMN\angle PMN, we can write it as:
tan45=PerpendicularBase=NPMN=mm\Rightarrow \tan 45^\circ = \dfrac{{Perpendicular}}{{Base}} = \dfrac{{NP}}{{MN}} = \dfrac{m}{m}
This can be simplified by dividing numerator and denominator by ‘m’
tan45=mm=1\Rightarrow \tan 45^\circ = \dfrac{m}{m} = 1
Therefore, we get the required value of tanπ4\tan \dfrac{\pi }{4} or tan45\tan 45^\circ as 11.

Note: In this question, we used an example of an equilateral triangle that was a crucial part of the solution. Notice that the value of tan45\tan 45^\circ does not depend on the sides of the triangle. An alternative approach to this problem can be to find the value of \sin 45^\circ {\text{ & }}\cos 45^\circ and then use the relation tanθ=sinθcosθ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} .