Question

If 4tanθ = 3, evaluate $\frac{4\sin \theta -\cos \theta +1}{4\sin \theta +\cos \theta -1}$

Answer 1

Given: 4tanθ = 3 ⇒ tan θ = $\frac{3}{4}$
Let us suppose a right angle triangle ABC right angled at B, with one of the acute angle θ. Let the sides be BC = 3k and AB = 4k, where k is a positive number.

By Pythagoras theorem, we get
$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{BC}}^2+{\mathrm{AB}}^2 \\ {\mathrm{AC}}^2={\left(3k\right)}^2+{\left(4k\right)}^2 \\ {\mathrm{AC}}^2=9k^2+16k^2 \\ \mathrm{AC}=\sqrt{25k^2} \\ \mathrm{AC}=\pm 5k \end{gathered}$$
Ignoring AC = − 5k , as k is a positive number, we get
AC = 5k
If $\tan \theta =\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3}{4}$, then $\sin \theta =\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{5}$ and $\cos \theta =\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4}{5}$
Putting the values in $\left(\frac{4\sin \theta -\cos \theta +1}{4\sin \theta +\cos \theta -1}\right)$, we get
$\left(\frac{4\times \frac{3}{5}-\frac{4}{5}+1}{4\times \frac{3}{5}+\frac{4}{5}-1}\right)=\left(\frac{{\displaystyle \frac{12-4+5}{5}}}{{\displaystyle \frac{12+4-5}{5}}}\right)=\frac{13}{11}$