If $$\cot \theta = \frac{3}{4},  then  \sin \theta + \cos \theta - \tan \theta$$ is equal to:

Given that,

$$\cot \theta = \frac{3}{4}$$
Then $$\tan \theta=\frac{4}{3}$$

We know for the right angled triangle,
$$Hypotenuse^2=Base^2+Perpendicular^2$$
So, $$ =\sqrt{3^2+4^2}$$
$$ =\sqrt{25}$$
$$ =5$$
Hence, $$\sin\theta=\dfrac{4}{5}$$ and $$\cos\theta=\dfrac{3}{5}$$ and

$$\Rightarrow \sin \theta + \cos \theta - \tan \theta$$
$$\Rightarrow \dfrac{4}{5}+ \dfrac{3}{5} - \dfrac{4}{3}$$
$$\Rightarrow \dfrac{4\times3}{5\times3}+ \dfrac{3\times3}{5\times3} - \dfrac{4\times5}{3\times5}$$
$$\Rightarrow \dfrac{12+9-20}{5\times3}$$
$$\Rightarrow \dfrac{1}{15}$$