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]
