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What is [tex]$\tan 45^{\circ}$[/tex]?

A. [tex]$\frac{1}{\sqrt{2}}$[/tex]
B. 1
C. [tex][tex]$\frac{1}{2}$[/tex][/tex]
D. [tex]$\sqrt{2}$[/tex]


Sagot :

To determine the value of [tex]\(\tan 45^{\circ}\)[/tex], we can start by recalling some fundamental trigonometric concepts. The tangent function, [tex]\(\tan \theta\)[/tex], is defined as the ratio of the sine of an angle to the cosine of that angle:

[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]

We will specifically evaluate this at [tex]\(\theta = 45^{\circ}\)[/tex].

Let's consider the angle [tex]\(45^{\circ}\)[/tex]:

1. In a right triangle where one of the angles is [tex]\(45^{\circ}\)[/tex], the other non-right angle is also [tex]\(45^{\circ}\)[/tex], making it an isosceles right triangle.
2. Thus, the two legs opposite these angles are equal in length. For simplicity, let's assume each leg has a length of 1 unit.
3. Using the Pythagorean theorem [tex]\(a^2 + b^2 = c^2\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the legs of the triangle and [tex]\(c\)[/tex] is the hypotenuse, we can calculate the hypotenuse as follows:
[tex]\[ 1^2 + 1^2 = c^2 \implies 1 + 1 = c^2 \implies c^2 = 2 \implies c = \sqrt{2} \][/tex]

Given that the triangle is isosceles with legs of 1 unit each, and the hypotenuse is [tex]\(\sqrt{2}\)[/tex], we can find the sine and cosine of [tex]\(45^{\circ}\)[/tex]:
[tex]\[ \sin 45^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}} \][/tex]
[tex]\[ \cos 45^{\circ} = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}} \][/tex]

Substituting these values into the tangent function:
[tex]\[ \tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1 \][/tex]

Thus, the value of [tex]\(\tan 45^{\circ}\)[/tex] is:
[tex]\[ \tan 45^{\circ} = 1 \][/tex]

Therefore, the correct answer is:
B. 1