Question

If 3 tan $\theta =4$, show that $\frac{(4cos\theta -sin\theta )}{(2cos\theta +sin\theta )}=\frac{4}{5}$

Answer 1

$3tan\theta =4tan\theta =\frac{4}{3}tan\theta =\frac{Oppositeside}{Adjacentside}Oppositeside=4Adjacentside=3Hypotenuse=\sqrt{(Oppositeside{)}^2+(Adjacentside{)}^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5Hypotenuse=5$

$sin\theta =\frac{Oppositeside}{Hypotenuseside}=\frac{4}{5}cos\theta =\frac{Adjacentside}{Hypotenuseside}=\frac{3}{5}$

$\frac{(4cos\theta -sin\theta )}{(2cos\theta +sin\theta )}=\frac{4\times \frac{3}{5}-\frac{4}{5}}{2\times \frac{3}{5}+\frac{4}{5}}=\frac{\frac{12}{5}-\frac{4}{5}}{\frac{6}{5}+\frac{4}{5}}=\frac{\frac{8}{5}}{\frac{10}{5}}=\frac{8}{10}=\frac{4}{5}$

Hence proved