Answer: \(12\) square units.

Answer: \(12\) square units.

For \(y=1+\tan3x\),

\(\displaystyle \frac{dy}{dx}=3\operatorname{sec}^{2} 3x\).

At \(x=\frac{\pi}{12}\),

\(3x=\frac{\pi}{4}\), so \(\tan3x=1\) and \(y=2\).

Thus \(P=\left(\frac{\pi}{12},2\right)\).

Also,

\(\displaystyle \frac{dy}{dx}=3\operatorname{sec}^{2}\frac{\pi}{4}=3(2)=6\).

The gradient of the normal is therefore \(-\frac16\).

The normal through \(P\) has equation

\(\displaystyle y-2=-\frac16\left(x-\frac{\pi}{12}\right)\).

At \(Q\), \(y=0\), so

\(\displaystyle -2=-\frac16\left(x-\frac{\pi}{12}\right)\).

Hence

\(\displaystyle x=12+\frac{\pi}{12}\).

The point \(R\) is \(\left(\frac{\pi}{12},0\right)\), so the horizontal length \(QR\) is 12 and the vertical height \(PR\) is 2.

Therefore the area of triangle \(PQR\) is

\(\displaystyle \frac12(12)(2)=12\).