Solve this

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).