If $$3 \sin \theta = 4 \cos \theta,  then  \tan^2 \theta + \sin \theta - \cos \theta$$ is equal to:

Given that,

$$3 \sin \theta = 4 \cos \theta$$

$$\tan \theta=\dfrac{4}{3}$$

Let AB=4 and BC=3

From the phythagoras theorem,

$$AB^2+BC^2=AC^2$$

Now, substituting the values,

$$\Rightarrow AC=\sqrt{AB^2+BC^2}$$

$$\Rightarrow AC=\sqrt{4^2+3^2}$$

$$\Rightarrow AC=\sqrt{25}=5$$

Now, substituting the values in  $$\tan^2 \theta + \sin \theta - \cos \theta$$

$$\Rightarrow \dfrac{4}{3}^2+\dfrac{4}{5}-\dfrac{3}{5}$$

$$\Rightarrow \dfrac{16}{9}+\dfrac{4-1}{5}$$

$$\Rightarrow \dfrac{16}{9}+\dfrac{1}{5}$$

$$\Rightarrow \dfrac{80}{45}+\dfrac{9}{45}$$

$$\Rightarrow \dfrac{89}{45}$$