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I'm having difficulty eliminating the parameter in the equations: $x = (tan^2\theta)$, $y = sec\theta$. The only strategy I know of for tackling trig parameters is to use the identity [$sin^2(x) + cos^2(x) = 1$] before setting that equal to some expression of $x + y$, but tangent gives me $x = \frac{sin^2\theta}{cos^2\theta}$, and I have no idea how to eliminate the denominator to get part of the identity. Am I just going about this completely wrong?

Thank you!

2 Answers 2

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Squaring $y = \sec \theta$

$y^2 = \sec^2 \theta$

We know that,

$\sec^2 \theta - \tan^2 \theta = 1$

So we have,

$y^2 - x = 1$

Other method to drive. As you said,

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

Divide above equation by $\cos^2 \theta$

$\tan^2 \theta + 1 = \sec^2 \theta$

$x + 1 = y^2$

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    Mine pleasure.. – 2017-01-27
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Use,

$$\sin^2(\theta)+\cos^2(\theta)=1$$

Dividing both sides by $\cos^2 (\theta)$ gives:

$$\tan ^2 (\theta)+1=\sec^2 (\theta)$$

This should be enough to conclude.

$$x+1=y^2$$

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    You are doing it completely wrong. – 2017-01-27
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    Why? @KanwaljitSingh – 2017-01-27
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    Uh...he did exactly the same thing you did, but with the $\tan^2$ on the other side of the equation. Saying "that's all wrong" without pointing out the location of the error is ... not very helpful, and against the spirit of the site. – 2017-01-27
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    Thank you both for all your help! I missed the identity you both used, that makes a lot more sense! – 2017-01-27
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    @John Hughes actually his complete statement is wrong. That's why I wrote just that. – 2017-01-27
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    By "his", do you mean Ahmed's, or JMartin's original? If the latter, you should comment on the original question rather than on Ahmed's answer. And in either case, you should be polite enough to point out specifically what is wrong. I see nothing wrong at all in Ahmed's answer. – 2017-01-27