Question

Given that x=1+2tan theta and y=3sec theta - 4, by eliminating theta, determine a relationship between x and y only !

Answer 1

$x=1+2\tan\theta\\\\ 2\tan\theta=x-1\\\\ \tan\theta=\frac{1}{2}(x-1)\\\\\\ y=3\sec\theta-4\\\\ 3\sec\theta=y+4\\\\ \sec\theta=\frac{1}{3}(y+4)$

We'll use the formula below

$\tan^2\alpha+1=\sec^2\alpha\iff \sec^2\alpha-\tan^2\alpha=1$

So:

$\sec^2\theta-\tan^2\theta=1\\\\ (\frac{1}{3}(y+4))^2-(\frac{1}{2}(x-1))^2=1\\\\ \boxed{\dfrac{(y+4)^2}{9}-\dfrac{(x-1)^2}{4}=1}$

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$x=\sin\theta\\\\\\ y=\cos^2\theta-3\\\\ \cos^2\theta=y+3$

We'll use the formula below

$\sin^2\alpha+\cos^2\alpha=1$

So:

$\sin^2\theta+\cos^2\theta=1\\\\ x^2+(y+3)=1\\\\ \boxed{x^2+y=-2}$