Question

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

Answer 1

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
We know that tan $\theta$ = $\frac{AB}{BC}$ = $\frac{4}{3}$

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AC² = 16k² + 9k²
⇒ AC² = 25k²
⇒ AC = 5k

Now, we have:

sin $\theta$ = $\frac{AB}{AC}=\frac{4k}{5k}=\frac{4}{5}$

cos $\theta$ = $\frac{BC}{AC}=\frac{3k}{5k}=\frac{3}{5}$

Substituting these values in the given expression, we get:

$$\begin{gathered} \frac{4\cos \theta -\sin \theta }{2\cos \theta +\sin \theta } \\ =\frac{4\left({\displaystyle \frac{3}{5}}\right)-{\displaystyle \frac{4}{5}}}{2\left({\displaystyle \frac{3}{5}}\right)+{\displaystyle \frac{4}{5}}} \\ =\frac{{\displaystyle \frac{12}{5}}-{\displaystyle \frac{4}{5}}}{{\displaystyle \frac{6}{5}}+{\displaystyle \frac{4}{5}}} \\ =\frac{{\displaystyle \frac{12-4}{5}}}{{\displaystyle \frac{6+4}{5}}} \\ =\frac{8}{10}=\frac{4}{5}=\mathrm{RHS} \end{gathered}$$
i.e., LHS = RHS

Hence proved.