$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$
$ \displaystyle \ \ \ \sin \theta =y$
$ \displaystyle \ \ \ \cos \theta =x$
$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$
$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$
$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$
$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$
$ \displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle OPN,$
$ \displaystyle \ \ \ {y}'=x\ \text{and }{x}'=y\ \text{numerically}\text{.}$
$ \displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the fourth quadrant}\text{.}$
$ \displaystyle \therefore {y}'=-x\ \text{and }{x}'=y.$
$ \displaystyle \ \ \ \sin (270{}^\circ +\theta )={y}'=-x=-\cos \theta $
$ \displaystyle \ \ \ \cos (270{}^\circ +\theta )={x}'=y=\sin \theta $
$ \displaystyle \ \ \ \tan (270{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{{x}}{{y}}=\cot \theta $
$ \displaystyle \ \ \ \cot (270{}^\circ +\theta )=\frac{{{x}'}}{{{y}'}}=-\frac{{y}}{{x}}=\tan \theta $
$ \displaystyle \ \ \ \sec (270{}^\circ +\theta )=\frac{1}{{{x}'}}=\frac{1}{y}=\operatorname{cosec}\theta $
$ \displaystyle \ \ \ \operatorname{cosec}(270{}^\circ +\theta )=\frac{1}{{{y}'}}=-\frac{1}{x}=-\sec\theta $
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