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2 tan 30°
II
1 + tan- 300​

Sagot :

MrRoyal

Question:

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex]

Answer:

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex][tex]= sin(60^{\circ})[/tex]

Step-by-step explanation:

Given

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex]

Required

Simplify

In trigonometry:

[tex]tan(30^{\circ}) = \frac{1}{\sqrt{3}}[/tex]

So, the expression becomes:

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}[/tex]

Simplify the denominator

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}[/tex]

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}[/tex]

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}[/tex]

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}[/tex]

Express the fraction as:

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex][tex]= \frac{2}{\sqrt 3} / \frac{4}{3}[/tex]

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{2}{\sqrt 3} * \frac{3}{4}[/tex]

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{1}{\sqrt 3} * \frac{3}{2}[/tex]

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{3}{2\sqrt 3}[/tex]

Rationalize

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}[/tex]

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{3\sqrt{3}}{2* 3}[/tex]

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex] [tex]= \frac{\sqrt{3}}{2}[/tex]

In trigonometry:

[tex]sin(60^{\circ}) = \frac{\sqrt{3}}{2}[/tex]

Hence:

[tex]\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}[/tex][tex]= sin(60^{\circ})[/tex]

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