# Thread: trig identities: eliminating θ

1. ## trig identities: eliminating θ

Hello. Thanks for any help in advance.

just having trouble with the method of eliminating θ from the following pairs of equations:
a. x = 2 cosθ, y = 2 sin θ

b. x = b tan, y = a sec θ

Hello. Thanks for any help in advance.

just having trouble with the method of eliminating θ from the following pairs of equations:
a. x = 2 cosθ, y = 2 sin θ

b. x = b tan, y = a sec θ
The key is to use the many forms of the Pythagorean Identity - in this case $\displaystyle \sin^2{\theta} + \cos^2{\theta} = 1$ and $\displaystyle \tan^2{\theta} + 1 = \sec^2{\theta}$.

(a) $\displaystyle x = 2\cos{\theta}, y = 2\sin{\theta}$

$\displaystyle x^2 = 4\cos^2{\theta}, y^2 = 4\sin^2{\theta}$

$\displaystyle x^2 + y^2 = 4\cos^2{\theta} + 4\sin^2{\theta}$

$\displaystyle x^2 + y^2 = 4(\cos^2{\theta} + \sin^2{\theta})$

$\displaystyle x^2 + y^2 = 4\cdot 1$

$\displaystyle x^2 + y^2 = 4$.

Can you follow a similar process for (b)?

3. for b. x = b tan θ, y = a sec θ would it be something like

(b tan θ)^2 + 1 = (a sec θ)^2

b^2 tan^2 θ + b^2 a^2 = a^2 sec^2 θ

b^2 (tan^2 θ + a^2)^2 = a^2 (sec^2 θ)

b^2 x^2 + a^2 b^2 = a^2 y^2

not too sure....=\

for b. x = b tan θ, y = a sec θ would it be something like

(b tan θ)^2 + 1 = (a sec θ)^2

b^2 tan^2 θ + b^2 a^2 = a^2 sec^2 θ

b^2 (tan^2 θ + a^2)^2 = a^2 (sec^2 θ)

b^2 x^2 + a^2 b^2 = a^2 y^2

not too sure....=\
You're on the right track. Remember that $\displaystyle \tan^2{\theta} + 1 = \sec^2{\theta}$, so $\displaystyle \sec^2{\theta} - \tan^2{\theta} = 1$.

If $\displaystyle x = b\tan{\theta}, y = a\sec{\theta}$

then $\displaystyle x^2 = b^2\tan^2{\theta}, y^2 = a^2\sec^2{\theta}$

$\displaystyle a^2x^2 = a^2b^2\tan^2{\theta}, b^2y^2 = a^2b^2\sec^2{\theta}$.

$\displaystyle a^2b^2\sec^2{\theta} - a^2b^2\tan^2{\theta} = b^2y^2 - a^2x^2$

$\displaystyle a^2b^2(\sec^2{\theta} - \tan^2{\theta}) = b^2y^2 - a^2x^2$

$\displaystyle a^2b^2\cdot 1 = b^2y^2 - a^2x^2$

$\displaystyle a^2b^2 = b^2y^2 - a^2x^2$

$\displaystyle 1 = \frac{b^2y^2 - a^2x^2}{a^2b^2}$

$\displaystyle \frac{b^2y^2}{a^2b^2} - \frac{a^2x^2}{a^2b^2} = 1$

$\displaystyle \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

5. ok thanks for your help.