2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $\cos^2 \theta$

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

$x + 1 = y^2$

$\endgroup$
  • $\begingroup$ Mine pleasure.. $\endgroup$ – Kanwaljit Singh Jan 27 '17 at 3:07
2
$\begingroup$

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.

$$x+1=y^2$$

$\endgroup$
  • $\begingroup$ You are doing it completely wrong. $\endgroup$ – Kanwaljit Singh Jan 27 '17 at 3:03
  • $\begingroup$ Why? @KanwaljitSingh $\endgroup$ – Ahmed S. Attaalla Jan 27 '17 at 3:04
  • 1
    $\begingroup$ 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. $\endgroup$ – John Hughes Jan 27 '17 at 3:05
  • $\begingroup$ Thank you both for all your help! I missed the identity you both used, that makes a lot more sense! $\endgroup$ – JMartin Jan 27 '17 at 3:06
  • $\begingroup$ @John Hughes actually his complete statement is wrong. That's why I wrote just that. $\endgroup$ – Kanwaljit Singh Jan 27 '17 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.