I have difficulty on how to eliminate parameter especially the equation involved trigonometry equation.
The question is asking for the area bounded by the curve , the 2 axes and the line $y=1$.
$x=4 \sin t$
$y=\cot t$
where $t$ is in the range $(0, \pi)$. I have tried to use the trigonometric formula $1+ \cot^2 t =\csc^2 t$
but I got the wrong answer.The answer given is 4ln[(square root 2)+1].
The Desmos plot of the relevant part of the curve looks like this:
The desired area is simply the area "under" (to the left of) the curve from $y=0$ to $y=1$.
We now express $x$ in terms of $y$. Since $1+\cot^2t=\csc^2t=\frac1{\sin^2t}$ and thus $\frac1{1+\cot^2t}=\sin^2t$, we have $$x=4\sin t=\frac4{\sqrt{1+\cot^2t}}=\frac4{\sqrt{1+y^2}}$$ So the desired area is $$\int_0^1\frac4{\sqrt{1+y^2}}\,dy=4[\sinh^{-1}y]_0^1=4\sinh^{-1}1=4\ln(1+\sqrt2)$$ The last equality is well-known; see e.g. OEIS A091648.