Question:
If $4 \cot \theta=3$ then $\left(\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}\right)=?$
(a) $\frac{3}{7}$
(b) $\frac{2}{7}$
(c) $\frac{1}{7}$
(d) 0
Solution:
Given: $4 \cot \theta=3$
$\Rightarrow \cot \theta=\frac{3}{4}$
Since, $\cot \theta=\frac{B}{P}$
$\Rightarrow P=4$ and $B=3$
Using Pythagoras theorem,
$P^{2}+B^{2}=H^{2}$
$\Rightarrow 4^{2}+3^{2}=H^{2}$
$\Rightarrow H^{2}=16+9$
$\Rightarrow H^{2}=25$
$\Rightarrow H=5$
Therefore,
$\sin \theta=\frac{P}{H}=\frac{4}{5}$
$\cos \theta=\frac{B}{H}=\frac{3}{5}$
$\left(\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}\right)=\left(\frac{\frac{4}{5}-\frac{3}{5}}{\frac{4}{5}+\frac{3}{5}}\right)$
$=\left(\frac{\frac{4-3}{5}}{\frac{4+3}{5}}\right)$
$=\frac{1}{7}$
Hence, the correct option is (c).