# If 5 tan θ − 4 = 0, then the value of 5 sin θ−4 cos θ5 sin θ+4 cos θ is

Question:

If $5 \tan \theta-4=0$, then the value of $\frac{5 \sin \theta-4 \cos \theta}{5 \sin \theta+4 \cos \theta}$ is

(a) $\frac{5}{3}$

(b) $\frac{5}{6}$

(c) 0

(d) $\frac{1}{6}$

Solution:

Given that $5 \tan \theta-4=0 .$ We have to find the value of the following expression

$\frac{5 \sin \theta-4 \cos \theta}{5 \sin \theta+4 \cos \theta}$

Since $5 \tan \theta-4=0 \Rightarrow \tan \theta=\frac{4}{5}$

We know that: $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

$\Rightarrow$ Base $=5$

$\Rightarrow$ Perpendicular $=4$

$\Rightarrow$ Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$

$\Rightarrow$ Hypotenuse $=\sqrt{16+25}$

$\Rightarrow$ Hypotenuse $=\sqrt{41}$

Since $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ and $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

Now we find

$\frac{5 \sin \theta-4 \cos \theta}{5 \sin \theta+4 \cos \theta}$

$=\frac{5 \times \frac{4}{\sqrt{41}}-4 \times \frac{5}{\sqrt{41}}}{5 \times \frac{4}{\sqrt{41}}+4 \times \frac{5}{\sqrt{41}}}$

$=\frac{\frac{20}{\sqrt{41}}-\frac{20}{\sqrt{41}}}{\frac{20}{\sqrt{41}}+\frac{20}{\sqrt{41}}}$

$=0$

Hence the correct option is $(c)$