$ = \frac{{n{{\sec }^n}\theta \tan \theta + n{{\cos }^{n - 1}}\theta \sin \theta }}{{\sec \theta \tan \theta + \sin \theta }}$
$ = \frac{{n({{\sec }^n}\theta + {{\cos }^n}\theta )}}{{\sec \theta + \cos \theta }}$ (Dividing ${N^r}$ and ${D^r}$ by $\tan \theta $)
==> ${\left( {\frac{{dy}}{{dx}}} \right)^2} = \frac{{{n^2}{{({{\sec }^n}\theta + {{\cos }^n}\theta )}^2}}}{{{{(\sec \theta + \cos \theta )}^2}}}$
$ = \frac{{{n^2}[{{({{\sec }^n}\theta - {{\cos }^n}\theta )}^2} + 4{{\sec }^n}\theta {{\cos }^n}\theta ]}}{{{{(\sec \theta - \cos \theta )}^2} + 4\sec \theta .\cos \theta }} = \frac{{{n^2}({y^2} + 4)}}{{{x^2} + 4}}$
==> $({x^2} + 4){\rm{ }}{\left( {\frac{{dy}}{{dx}}} \right)^2} = {n^2}({y^2} + 4)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$7 x+11 y+\alpha z=13$
$5 x+4 y+7 z=\beta$
$175 x+194 y+57 z=361$
has infinitely many solutions, then $\alpha+\beta+2$ is equal to
$R _{1}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \in Q \right\}$ and $R _{2}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \notin Q \right\}$
where $Q$ is the set of all rational numbers. Then