Question

How do you solve ${\tan }^2\theta =1$?

Answer 1

$\theta =\pm \frac{\pi }{8}$
Explanation :

If you want a certain quantity to equal $1$ when squared, then your quantity must be either $1$ or $-1$. Every other number would not equal $1$ when squared.

Since our quantity is $\tan \left(\theta \right)$, we're asking for $\tan \left(\theta \right)=1$ or $\tan \left(\theta \right)=-1$

You could either look for a table of known values to

$\tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}$
So $\tan \left(\theta \right)=1$ leads to

$\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}=1\iff \sin \left(\theta \right)=\cos \left(\theta \right)$

and the sine and cosine functions have the same value where $\theta ={45}^{\circ }=\frac{\pi }{4}$ radians.

For the same reason. $\tan \left(\theta \right)=-1$ leads to $\sin \left(\theta \right)=-\cos \left(\theta \right)$ which happens for $\theta =-{45}^{\circ }$ radians. You can find this second solution remembering that the tangent is an odd function, i.e. $\tan (-x)=-\tan \left(x\right)$