# Exact value of $\tan^2 \alpha -\cot^2 \alpha$?

Knowing that $\alpha$ is acute and that $\sin \alpha-\cos \alpha = \dfrac{1}{3}$, determine the exact value of

$$\tan^2 \alpha -\cot^2 \alpha.$$

I've used $\tan\alpha= \dfrac{\sin\alpha}{\cos\alpha}$ but can't seem to get anywhere. I get stuck here:

$$\dfrac{\sin \alpha + \cos \alpha}{3\cos^2 \alpha \sin^2\alpha}$$

when I use $\sin^2 \alpha + \cos^2 \alpha = 1$ here it seems to get even more complicated...

Squaring $\sin\alpha-\cos\alpha = \frac13$, we get $$\sin^2\alpha - 2\sin\alpha\cos\alpha + \cos^2\alpha = \frac19\\ \sin\alpha\cos\alpha = \frac{4}{9}$$ which helps you with the denominator of your fraction.

We also get from this that $$\frac{17}{9} = 1 + 2\sin\alpha\cos\alpha \\= \sin^2\alpha + 2\sin\alpha\cos\alpha + \cos^2\alpha \\= (\sin\alpha + \cos\alpha)^2$$ Knowing that $\alpha$ is acute, you can take the square root here, and that should give you the numerator of your fraction.

Let $, c=cos\alpha, s=\sin\alpha$

$tan^2\alpha-cot^2\alpha=\frac{(s-c)(s+c)}{(cs)^2}$.

Let $s+c=k\Longrightarrow (s+c)^2=s^2+2sc+c^2=1+2sc=\frac{17}{9}=k^2\Longrightarrow k=-+\frac{\sqrt{17}}{3}$.

Hence we know $(s-c)$, $(s+c)$ and $sc$. By putting them above equality we can obtain result.

But, $tan^2\alpha-cot^2\alpha$ can take $2$ value. Maybe, I missed something.

(sin A-cos A)^2=1/9 sin A^2 + cos A^2 -2sin Acos A = 1/9 1+sin 2A = 1/9 so expand sin 2A in tanA then you can get a value to tanA then easily you can come to your answer (I don't know how to type alpha in this thing instead I used A)