Trigonometric equation: $2 \cot^2 \alpha - 2 \sec^2 \alpha = 1$

Question

Can someone help me solving this equation:

$$ 2 \cot^2 \alpha - 2 \sec^2 \alpha = 1 $$

I have to solve this for $\alpha$. Is there any relation I'm missing? I guess I'm going to have to bring both to the same trigonometric number?

Answer 1

Hint

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

so

$$\frac{1}{\tan^2\alpha}-1-\tan^2\alpha=\frac{1}{2}$$

Now use $u=\tan^2\alpha$

$$\frac{1}{u}-u-\frac{3}{2}=0 \rightarrow 2u^2+3u-2=0$$

What give us $u=-2$ or $u=1/2$.

Can you finish?

Answer 2

we have by definition $$2\left(\frac{\cos(x)^2}{\sin(x)^2}-\frac{1}{\cos(x)^2}\right)=1$$ this gives $$2(\cos(x)^2-\sin(x)^2)=\sin(x)^2\cos(x)^2$$ with $$\sin(x)^2=1-\cos(x)^2$$ we obtain $$2(\cos(x)^4-(1-\cos(x)^2)=(1-\cos(x)^2)\cos(x)^2$$ this is equivalent to $$3\cos(x)^4+\cos(x)^2-2=0$$

Answer 3

$$ \sec^2(\alpha)=1+\tan^2(\alpha) $$ And $$ \cot^2(\alpha)=\frac{1}{\tan^2(\alpha)} $$ Simplifying we get a quadratic equation which can solved to get the value of $\tan^2(\alpha)$