Question:

1. Given condition
\[3\cot A = 4 \quad\Rightarrow\quad \cot A = \frac{4}{3}.
\]
Since \(\cot A = \frac{1}{\tan A}\), we have
\[\tan A = \frac{3}{4}.
\]

2. Compute the left‑hand side (LHS)
\[\text{LHS}=\frac{2\tan A}{1+\tan^{2} A}.
\]
Substituting \(\tan A = \frac{3}{4}\):
\[\tan^{2} A = \left(\frac{3}{4}\right)^{2}=\frac{9}{16},\qquad
1+\tan^{2} A = 1+\frac{9}{16}=\frac{25}{16}.
\]
Numerator:
\[2\tan A = 2\times\frac{3}{4}=\frac{3}{2}.
\]
Hence
\[\text{LHS}=\frac{\frac{3}{2}}{\frac{25}{16}}=\frac{3}{2}\times\frac{16}{25}=\frac{48}{50}=\frac{24}{25}.
\]

3. Find \(\sin A\) and \(\cos A\)
From \(\tan A = \frac{3}{4}\), consider a right‑angled triangle with opposite side = 3, adjacent side = 4. Then the hypotenuse is \(\sqrt{3^{2}+4^{2}}=5\).
Therefore
\[\sin A = \frac{3}{5},\qquad \cos A = \frac{4}{5}.
\]

4. Compute the right‑hand side (RHS)
Using double‑angle formulas:
\[\sin 2A = 2\sin A\cos A = 2\times\frac{3}{5}\times\frac{4}{5}=\frac{24}{25},\]
\[\cos 2A = \cos^{2} A-\sin^{2} A = \left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}.
\]
Hence
\[\text{RHS}=\cos 2A-\sin 2A = \frac{7}{25}-\frac{24}{25}= -\frac{17}{25}.
\]

5. Comparison
\[\text{LHS}=\frac{24}{25},\qquad \text{RHS}= -\frac{17}{25}.
\]
Since \(\frac{24}{25}
eq -\frac{17}{25}\), the given equality does not hold.

6. Conclusion
The statement \(\displaystyle \frac{2\tan A}{1+\tan^{2} A}=\cos 2A-\sin 2A\) is false for the angle satisfying \(3\cot A =4\).