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How do I find the solutions of this equation:
$$\tan^2 (x)=-1$$

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4 Answers 4

There is no solution since any real number squared is always positive.

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(Edited after Julian's comment)

There might be complex solutions. In order to find these write

$$\tan z={1\over i}{e^{iz}-e^{-iz}\over e^{iz}+e^{-iz}}={1\over i}{u-1\over u+1}$$

with $u:=e^{2iz}$. With these conventions we have to solve the equation

$$-\left({u-1\over u+1}\right)^2=-1\ ,\quad{\rm i.e.}\quad {u-1\over u+1}=\pm 1\ .$$

As $(u-1)/(u+1)$ cannot be $1$ the last equation has the single solution $u=0$, so that we now have to determine the solutions $z\in{\mathbb C}$ of the equation

$$e^{2iz}\ = 0\ ;$$

but there is no such $z$.

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There is a $1/i$ missing in the formula for $\tan z$. This makes a great difference, since there are no solutions, real or complex. This can be seen from Chandrasekhar's answer, since $\sec x=0$ has no solution, real or complex. –  Julián Aguirre Dec 18 '11 at 12:44
@Julian Aguirre: How could I! –  Christian Blatter Dec 18 '11 at 14:44
Thank you for your responses! could you help me in solving this following equation too,$$tanx=2x/(1-x^2 )$$ –  alice Dec 18 '11 at 14:58

Hint: $$ \tan^{2}(x) = -1 \Rightarrow 1 + \tan^{2}(x) = 0 \Rightarrow \sec^{2}(x) = 0.$$

Oh, on the other hand note that $x^{2} = -1$ has no real solutions.

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Notice that $\tan^2(0) = 0$ and for any other $x \neq 0$, $\tan^2(x) > 0$.

If you plot $\tan^2(x)$ and $-1$ you will see that those 2 curves never intersect, because $\tan^2(x) \geq 0$, $x\in{\mathbb R}$.

enter image description here

Therefore there are no real solutions.

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