Please, help toconvert the Polar Equation to Cartesian Coordinates: r^{2}=\sec4\theta

Let P(x, y) be the terminal point on the unit circle determined by t. Then $\sin t=,\cos t=,\text{and}\tan t=$.

Reaching upon 9=1 while solving x for ${\displaystyle 3\tan (x-{15}^{\circ })=\tan (x+{15}^{\circ })}$
Substituting ${\displaystyle y=x+{45}^{\circ }}$, we get
${\displaystyle 3\tan (y-{60}^{\circ })=\tan (y-{30}^{\circ })}$
${\displaystyle 3\frac{\tan y-\sqrt{3}}{1+\sqrt{3}\tan y}=\frac{\tan y-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}\cdot \tan y}}$

$3({\tan }^{2}y-3)=3{\tan }^2-1$

9=1

The solution provided by the book $x=n\pi +\frac{\pi }{4}$ fits, so why did i get 9=1?

How do I solve this inequality? ${\displaystyle 3\sin \left(2x\right)>\sin \left(x\right)+\cos \left(x\right)+1}$

If
${\displaystyle \frac{3-{\tan }^2\frac{\pi }{7}}{1-{\tan }^2\frac{\pi }{7}}=\alpha \cos \frac{\pi }{7}}$

Maximizing the sum ${\displaystyle \sum _{n=1}^m\sin n}$
For which value of m, we will obtain the maximum sum?
Here's my approach : ${\displaystyle \sum _{n=1}^m\sin n=\frac{\sin 1}{4{\sin }^2\frac{1}{2}}-\frac{2\cos (m+\frac{1}{2})}{4\sin \left(\frac{1}{2}\right)}}$
If we can minimize ${\displaystyle 2\cos (m+\frac{1}{2})}$ then this will result in maximizing the sum.
But the problem is I can't quite figure out what the minimum value of ${\displaystyle \cos (m+\frac{1}{2})}$ is.

I'm trying find an exact value for
${\displaystyle \cos \left(\frac{1}{3}\arctan \left(\frac{-10}{9\sqrt{3}}\right)\right)}$

If ${\displaystyle u=\sqrt{a{\cos }^{2}x+b{\sin }^{2}x}+\sqrt{b{\cos }^{2}x+a{\sin }^{2}x}}$, find the maximum and minimum value of ${\displaystyle u^2}$.