$ \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 PON,$
$ \displaystyle \ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}$
$ \displaystyle \begin{array}{l}\ \ \ \text{But it is clear that, and }{y}'\ \text{have }y\ \text{opposite signs}\\\ \ \text{ }{x}'\ \text{and }x\ \text{have the same sign,}\end{array}$
$ \displaystyle \therefore {y}'=-y\ \text{and }{x}'=x.$
$ \displaystyle \therefore \sin (-\theta )={y}'=-y=-\sin \theta $
$ \displaystyle \ \ \ \cos (-\theta )={x}'=x=\cos \theta $
$ \displaystyle \ \ \ \tan (-\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{y}{x}=-\tan \theta $
$ \displaystyle \ \ \ \cot (-\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=-\cot \theta $
$ \displaystyle \ \ \ \sec (-\theta )=\frac{1}{{{x}'}}=\frac{1}{x}=\sec \theta $
$ \displaystyle \ \ \ \operatorname{cosec}(-\theta )=\frac{1}{{{y}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta $
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