Question

Tg^2 2x-tg^2 x/1-tg^2 2x tg^2 x=tg 3x tg x

Answer 1

[(tg2x+tgx)/(1-tg2xtgx)]*[(tg2x-tgx)/(1+tg2xtgx)]=tg3x*tgx
tg3xtgx=tg3xtgx

Answer 2

To simplify the expression $\frac{\tan^2(2x) - \tan^2(x)}{1 - \tan^2(2x) \tan^2(x)}$ and prove that it's equal to $\tan(3x) \tan(x)$, you can use trigonometric identities. One identity that you'll need is the difference of squares identity for tangent:

$\tan^2(A) - \tan^2(B) = \tan(A + B) \tan(A - B)$

Let's apply this identity to your expression:

$\frac{\tan^2(2x) - \tan^2(x)}{1 - \tan^2(2x) \tan^2(x)} = \frac{\tan(2x + x) \tan(2x - x)}{1 - \tan(2x) \tan(x) \tan(2x) \tan(x)}$

Now, we can use another trigonometric identity:

$\tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A) \tan(B)}$

Applying this identity to the denominator, we get:

$1 - \tan(2x) \tan(x) \tan(2x) \tan(x) = 1 - \frac{\tan(2x) \tan(x) - \tan(2x) \tan(x)}{1 + \tan(2x) \tan(x) \tan(2x) \tan(x)}$

Now, let's substitute this back into the expression:

$\frac{\tan(2x + x) \tan(2x - x)}{1 - \frac{\tan(2x) \tan(x) - \tan(2x) \tan(x)}{1 + \tan(2x) \tan(x) \tan(2x) \tan(x)}}$

Simplify the numerator:

$\tan(3x) \tan(x)$

Now, we can simplify the expression:

$\frac{\tan(3x) \tan(x)}{1 - \frac{\tan(2x) \tan(x) - \tan(2x) \tan(x)}{1 + \tan(2x) \tan(x) \tan(2x) \tan(x)}}$

Let's simplify the denominator:

$1 - \frac{\tan(2x) \tan(x) - \tan(2x) \tan(x)}{1 + \tan(2x) \tan(x) \tan(2x) \tan(x)} = 1 - \frac{0}{1 + \tan^2(2x) \tan^2(x)} = 1$

So, the simplified expression is:

$\frac{\tan(3x) \tan(x)}{1} = \tan(3x) \tan(x)$

Thus, the expression $\frac{\tan^2(2x) - \tan^2(x)}{1 - \tan^2(2x) \tan^2(x)}$ is indeed equal to $\tan(3x) \tan(x)$.