We have been given that [tex]3\tan(\alpha)=4[/tex]. We are asked to prove that [tex]\frac{5\sin(\alpha)-3\cos(\alpha)}{3\sin(\alpha)+3\cos(\alpha)}=\frac{11}{21}[/tex].
First of all, we will solve for tangent of alpha as:
[tex]\tan(\alpha)=\frac{4}{3}[/tex]
Now, we will divide left side of our given equation by [tex]\cos(\alpha)[/tex] as:
[tex]\frac{\frac{5\sin(\alpha)}{\cos(\alpha)}-\frac{3\cos(\alpha)}{\cos(\alpha)}}{\frac{3\sin(\alpha)}{\cos(\alpha)}+\frac{3\cos(\alpha)}{\cos(\alpha)}}[/tex]
We know that [tex]\frac{\sin(x)}{\cos(x)}=\tan(x)[/tex], so we will get:
[tex]\frac{5\tan(\alpha)-3}{3\tan(\alpha)+3}[/tex]
Upon substituting [tex]\tan(\alpha)=\frac{4}{3}[/tex], we will get:
[tex]\frac{5\tan(\alpha)-3}{3\tan(\alpha)+3}=\frac{5\cdot\frac{4}{3}-3}{3\cdot\frac{4}{3}+3}[/tex]
[tex]\frac{5\tan(\alpha)-3}{3\tan(\alpha)+3}=\frac{\frac{20}{3}-3}{\frac{12}{3}+3}[/tex]
[tex]\frac{5\tan(\alpha)-3}{3\tan(\alpha)+3}=\frac{\frac{20}{3}-\frac{9}{3}}{\frac{12}{3}+\frac{9}{3}}[/tex]
[tex]\frac{5\tan(\alpha)-3}{3\tan(\alpha)+3}=\frac{\frac{20-9}{3}}{\frac{12+9}{3}}[/tex]
[tex]\frac{5\tan(\alpha)-3}{3\tan(\alpha)+3}=\frac{\frac{11}{3}}{\frac{21}{3}}[/tex]
[tex]\frac{5\tan(\alpha)-3}{3\tan(\alpha)+3}=\frac{11\cdot3}{21\cdot 3}[/tex]
[tex]\frac{5\tan(\alpha)-3}{3\tan(\alpha)+3}=\frac{11}{21}[/tex]
Hence proved.
