# Parameters with trig functions

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!

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 $\cos^2 \theta$

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

$x + 1 = y^2$

• Mine pleasure.. – Kanwaljit Singh Jan 27 '17 at 3:07

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.

$$x+1=y^2$$

• You are doing it completely wrong. – Kanwaljit Singh Jan 27 '17 at 3:03
• Why? @KanwaljitSingh – Ahmed S. Attaalla Jan 27 '17 at 3:04
• 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. – John Hughes Jan 27 '17 at 3:05
• Thank you both for all your help! I missed the identity you both used, that makes a lot more sense! – JMartin Jan 27 '17 at 3:06
• @John Hughes actually his complete statement is wrong. That's why I wrote just that. – Kanwaljit Singh Jan 27 '17 at 3:22