Given
[tex]\tan3x=\frac{\sin6x}{1+\cos6x}[/tex]
Find
Find which identities used to show
[tex]\tan3x=\frac{\sin6x}{1+\cos6x}[/tex]
Explanation
as,, we use
[tex]\begin{gathered} \sin2x=2\sin x\cos x \\ \cos2x=2\cos^2x-1 \end{gathered}[/tex]
so ,
[tex]\begin{gathered} \sin(6x)=2\sin(3x)\cos(3x) \\ \cos(6x)=2\cos^2(3x)-1 \end{gathered}[/tex]
now, substitute the values on the left side
[tex]\begin{gathered} \frac{2\sin(3x)\cos(3x)}{1+(2\cos^2(3x)-1)} \\ \\ \frac{\begin{equation*}2\sin(3x)\cos(3x)\end{equation*}}{1+2\cos^2(3x)-1} \\ \\ \frac{2\sin(3x)\cos(3x)}{2\cos^2(3x)} \\ \\ \frac{\sin(3x)}{\cos(3x)} \\ \\ \tan(3x) \end{gathered}[/tex]
Final Answer
Therefore . we use 3 identities which are A , B and D
