\(
x=1+2\tan\theta\\\\
2\tan\theta=x-1\\\\
\tan\theta=\frac{1}{2}(x-1)\\\\\\
y=3\sec\theta-4\\\\
3\sec\theta=y+4\\\\
\sec\theta=\frac{1}{3}(y+4)\)
We'll use the formula below
\(
\tan^2\alpha+1=\sec^2\alpha\iff \sec^2\alpha-\tan^2\alpha=1\)
So:
\(
\sec^2\theta-\tan^2\theta=1\\\\
(\frac{1}{3}(y+4))^2-(\frac{1}{2}(x-1))^2=1\\\\
\boxed{\dfrac{(y+4)^2}{9}-\dfrac{(x-1)^2}{4}=1}
\)
-----------------//----------------
\(
x=\sin\theta\\\\\\
y=\cos^2\theta-3\\\\
\cos^2\theta=y+3\)
We'll use the formula below
\(
\sin^2\alpha+\cos^2\alpha=1\)
So:
\(
\sin^2\theta+\cos^2\theta=1\\\\
x^2+(y+3)=1\\\\
\boxed{x^2+y=-2}
\)
