Solve the given Alternate Coordinate Systems and give a correct answer10) Convert the equation from Cartesian to polar coordinates solving for r^2: frac{x^2}{9} - frac{y^2}{16} = 25

Solution: Relation between polar Co-ordinate system and Cartesian Co-ordinate system ${\displaystyle x=r\cos 0y=r\sin 0}$

$r=\sqrt{x^2+y^2}$

given that - ${\displaystyle \frac{x^2}{9}-\frac{y^2}{16}=25}$ put $x=r\cos 0$, $y=e\sin 0$
$\Rightarrow \frac{r^{2}co{s}^{2}0}{0}-\frac{r^{2}si{n}^{2}0}{16}=25$
$\Rightarrow r^2[\frac{16co{s}^{2}0-9si{n}^{2}0}{177}]=25$
$\Rightarrow r^2=\frac{25\times 177}{[(7cos{)}^2-(3sin{)}^2]}[a^2-h^2=(a-b)(a+b)]$
$\Rightarrow r^2=\frac{25\times 177}{(7cos0-3sin0)(4cos0+3sin0)}$
$\Rightarrow r^2=\frac{25\times 177}{5[(\frac{7}{5}cos0-\frac{3}{5}sin0)]5[(\frac{7}{5}cos0+\frac{3}{5}sin0)]}$
$\Rightarrow r^2\frac{177}{(\frac{7}{5}cos0-\frac{3}{5}sin0)(\frac{7}{5}cos0+\frac{3}{5}sin0)}[sin({53}^{\circ })=\frac{3}{5}]$
$\Rightarrow r^2=\frac{177}{(sin{53}^{\circ }cos0-cos{53}^{\circ }sin0)(sin{53}^{\circ }cos0+cos{53}^{\circ }sin0)}$  formula 

${\displaystyle \sin (A+B)=\sin A\cos B+\cos A\sin B}$

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

$\Rightarrow r^2\frac{177}{\left[\sin (53-0)\right]\left[\sin (53+0)\right]}$

$\Rightarrow r^2=\frac{177}{\left[\sin (53-0)\right]\left[\sin (0+53)\right]}$

Finally answer is ${\displaystyle \Rightarrow r^2=\frac{177}{\left[\sin (53-0)\right]\left[\sin (0+53)\right]}}$