To find \cos\theta, we will use the Pythagorean identity :

\cos^2\theta + \sin^2\theta = 1

Given that \sin\theta = \frac{3}{5}, we substitute this value into the identity:

\cos^2\theta + \left(\frac{3}{5}\right)^2 = 1

\cos^2\theta + \frac{9}{25} = 1

Subtract \frac{9}{25} from both sides:

\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}

Taking the square root of both sides, we find:

\cos\theta = \pm \frac{4}{5}

Since \frac{3\pi}{4} < \theta < \frac{5\pi}{4}, \theta is in the third quadrant where cosine is negative . Therefore, \cos\theta = -\frac{4}{5} .

Now, we can find \tan\theta using the identity:

\tan\theta = \frac{\sin\theta}{\cos\theta}

Substitute the known values:

\tan\theta = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}

Next, we use the identity proven in part (c):

\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}

Substitute \tan A = \tan\theta = -\frac{3}{4} :

\tan 3\theta = \frac{3\left(-\frac{3}{4}\right) - \left(-\frac{3}{4}\right)^3}{1 - 3\left(-\frac{3}{4}\right)^2}

Substitute these into the expression for \tan 3\theta:

\tan 3\theta = \frac{-\frac{9}{4} + \frac{27}{64}}{-\frac{11}{16}}

To simplify, multiply the numerator and the denominator by 64:

\tan 3\theta = \frac{-\frac{9}{4} \cdot 64 + \frac{27}{64} \cdot 64}{-\frac{11}{16} \cdot 64}

This simplifies to:

\tan 3\theta = \frac{-144 + 27}{-44}

\tan 3\theta = \frac{-117}{-44}

\tan 3\theta = \frac{117}{44}