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**RogueDemon** 14. a) Which equations are not identities? Justify your answers.

b) For those equations that are identities, state any restrictions on the variables.

i) $\displaystyle (1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}$

__Left Side__

$\displaystyle = (1 - cos^2x)(1 - tan^2x)$

$\displaystyle = 1 - tan^2x - cos^2x + (-cos^2x)(-tan^2x)$

$\displaystyle = 1 - cos^2x - tan^2x + (cos^2x)(tan^2x)$

$\displaystyle = sin^2x - tan^2x + sin^2x$

$\displaystyle = (cos^2x)(tan^2x) - tan^2x + (cos^2x)(tan^2x)$

$\displaystyle = tan^2x(cos^2x - 1 + cos^2x)$

$\displaystyle = tan^2x(-sin^2x + cos^2x)$ Edit: Was an error made here?

$\displaystyle = tan^2x(-1)$

$\displaystyle = -tan^2x$

__Right Side:__

$\displaystyle = \frac{sin^2x - 2sin^4x}{1 - sin^2x}$

$\displaystyle = \frac{sin^2x - 2(sin^2x)(sin^2x)}{cos^2x}$

$\displaystyle = \frac{sin^2x(1 -2(1)(1))}{cos^2x}$

$\displaystyle = \frac{sin^2x(1 -2)}{cos^2x}$

$\displaystyle = \frac{sin^2x(-1)}{cos^2x}$

$\displaystyle = -tan^2x$

Left Side = Right Side

Therefore, this equation is an identity.

When I showed this solution to my teacher, I was told that had I used identities that we haven't learned yet. She said that on the exam, we were to only be able to use reciprocal identities and the Pythagorean identity $\displaystyle sin^2x + cos^2x = 1.$ She also told me that using these identities, I can make the $\displaystyle LS$ and $\displaystyle RS$ equal however with different values. Is there another way of proving this equation to be an identity?

Also, how do I know whether or not an equation will eventually end up as $\displaystyle LS = RS$? The only reason I know it with the few assigned questions is because there is an answer at the back of the book. Though on an exam I won't have access to these answers. Is it just trial and error? Because if it is, I can be going on for a pretty long time just trying to solve one question.