$3\dfrac{\text{sin 47°}}{\text{cos 43°}} - \dfrac{\text{tan 50°}}{\text{cot 40°}} - \text{2 sin}^2 45°$ is equal to :

1. 0

2. -1

3. 1

4. 2

Solving,

$\Rightarrow 3\dfrac{\text{sin 47°}}{\text{cos 43°}} - \dfrac{\text{tan 50°}}{\text{cot 40°}} - \text{2 sin}^2 45° \\[1em] \Rightarrow 3\dfrac{\text{sin 47°}}{\text{cos (90° - 47°)}} - \dfrac{\text{tan 50°}}{\text{cot (90° - 50°)}} - \text{2 sin}^2 45° \\[1em] \Rightarrow 3\dfrac{\text{sin 47°}}{\text{sin 47°}} - \dfrac{\text{tan 50°}}{\text{tan 50°}} - 2 \times \Big(\dfrac{1}{\sqrt{2}}\Big)^2 \\[1em] \Rightarrow 3 \times 1 - 1 - 2 \times \dfrac{1}{2} \\[1em] \Rightarrow 3 - 1 - 1 \\[1em] \Rightarrow 1.$

Hence, Option 3 is the correct option.